GfFl 


ELEMENTARY   PRACTICAL  MECHANICS 


ELEMENTARY 


PRACTICAL  MECHANICS 


BY 


J.  M.  JAMESON 

il 

Head  of  the  Department  of  Physics,  School  of  Science  and  Technology 
Pratt  Institute,  Brooklyn.  N.  Y. 


SECOND  EDITION 


LONGMANS,    GREEN,    AND    CO, 

FOURTH  AVENUE  &  30TH  STREET,  NEW  YORK 

LONDON,  BOMBAY,  AND  CALCUTTA 

1912 


Copyright,  1909 

BY 
LONGMANS,    GREEN,    AND   CO. 

First  published  September,  1910 

Second  Edition  December,  1910 

August,  1912 


THE  SCIENTIFIC  PRESS 
ROBERT   DRUMMOND   AND  COMPANY 

BROOKLYN,   N.  Y. 


PREFACE 


THE  accompanying  text  is  an  attempt  to  express  prac- 
tical mechanics  as  the  science  of  the  processes  and  struc- 
tures of  every-day  life  rather  than  as  a  series  of  more 
or  less  abstract  mathematical  demonstrations.  It  aims 
to  convey  not  only  a  knowledge  of  facts  and  of  funda- 
mental theory,  but  also  some  training  in  ability  to  apply 
such  knowledge;  to  equip  the  student  who  is  to  receive 
no  further  formal  instruction  in  mechanics  with  practical 
information  required  for  business  or  industrial  life,  and  to 
establish  a  foundation  and  right  point  of  view  for  the 
student  who  is  to  continue  his  study  along  applied  lines, 
which  shall  enable  him  to  advance  without  confusion  or 
loss  of  time  in  acquiring  new  methods  and  a  new  vocab- 
ulary. 

It  is  designed  primarily  as  a  text  for  elementary  tech- 
nical and  manual  training  schools  which  find  the  usual 
texts  in  elementary  mechanics  too  theoretical  or  too 
mathematical  for  their  needs,  and  which  require  a  more 
complete  and  practical  course  than  that  furnished  by  a 
text  in  general  physics;  also  as  an  introductory  text  for 
engineering  schools  to  be  followed  by  the  more  applied 
courses  in  mechanism,  and  in  the  electrical  and  mechanical 
laboratories.  The  subjects  of  work,  friction,  and  power 
transmission,  and  of  power  brake  and  dynamometer  tests 
are  discussed  with  considerable  detail,  and  a  chapter  is 


Vi  PREFACE 

devoted  to  elasticity  and  to  stress  in  materials.  If  properly 
supplemented  by  laboratory  exercises,  therefore,  it  is 
believed  that  satisfactory  material  will  be  found  for  a 
short  course  in  applied  mechanics. 

It  should,  perhaps,  be  stated  that  the  present  text  has 
grown  out  of  a  series  of  notes  issued  for  several  years 
in  mimeograph  form  to  the  students  in  the  School  of 
Science  and  Technology,  Pratt  Institute.  Nothing  is 
included,  therefore,  which  has  not  stood  the  test  of  several 
years7  use  with  students  in  elementary  technical  courses. 
The  time  given  to  the  subject  at  Pratt  Institute  is  repre- 
sented by  one-half  year  with  five  lectures  or  recitations 
and  six  hours  laboratory  per  week.  The  requirements  in 
mathematics  are  limited  to  simultaneous  and  simple  quad- 
ratic equations  and  the  simplest  trigonometric  functions. 

The  order  of  presentation  is  that  which  has  been  found 
to  give  best  results.  General  principles  and  definitions  to 
form  a  ground  work  for  laboratory  exercises  are  introduced 
early,  and  the  complete  statement  of  theory,  together  with 
its  applications,  is  then  reached  through  laboratory  and 
lecture  work  combined.  The  importance  of  carefully 
adapted  laboratory  exercises  in  teaching  mechanics  cannot 
be  over  emphasized.  Numerous  suggestions  of  models, 
etc.,  used  in  the  laboratories  at  Pratt  Institute  are  given 
in  the  figures  illustrating  this  text. 

In  statics,  as  a  rule,  both  the  graphical  and  the  analytical 
solutions  are  indicated.  If  desired,  therefore,  the  use  of 
trigonometric  functions  may  be  avoided  without  greatly 
modifying  the  ground  covered  in  the  text.  Special  effort 
has  been  made  to  present  in  a  clear  and  usable  form 
those  portions  of  the  text  which  deal  with  the  mechanics 
of  moving  bodies,  and  to  make  the  students7  work  in  this 
subject  something  more  than  blind  substitution  of  values 
in  an  uncomprehended  formula.  Repeated  drill  is  pro- 


PREFACE  vii 

vided  in  the  application  of  the  fundamental  principle 
f=ma  to  familiar  instances,  such  as  the  starting  and 
stopping  of  trains,  the  tension  in  hoisting  ropes,  etc.,  and 
of  the  corresponding  expression  for  torque  in  rotary  motion, 
to  shafting,  fly-wheels,  etc.  The  conception  of  moment 
of  inertia  and  radius  of  gyration  is  approached  through 
the  familiar  ideas  of  action  and  reaction  and  of  moment 
of  force  rather  than  through  more  abstract  mathematical 
reasoning. 

No  apology  is  offered  for  the  large  amount  of  space 
devoted  to  problems.  Too  many  problems  in  mechanics  are 
scarcely  possible.  It  is  believed  that  the  problems  in  this 
text  will  be  found  carefully  selected  and  well  distributed. 

It  is  our  practice  to  require  the  student  to  construct 
a  simple,  preliminary  diagram  of  force  conditions  as  a 
help  to  clear  thinking,  and  as  a  basis  for  the  final  mathex 
matical  statement  of  his  solution.  Diagrams  of  this  charac- 
ter have  been  inserted  freely  in  the  text,  and  an  attempt 
has  also  been  made  to  supply  a  number  of  cuts  from  pho- 
tographs of  actual  commercial  structures  which,  upon  the 
specification  of  the  necessary  angles  and  dimensions  by 
the  instructor,  shall  supply  material  for  valuable  problems. 
In  this  connection,  grateful  acknowledgment  is  returned 
to  Messrs.  John  Wiley  &  Sons,  publishers,  and  to  the 
respective  authors,  for  permission  to  reproduce  the  fol- 
lowing cuts:  Fig.  91  taken  from  Sanborn's  " Problems  in 
Mechanics;"  Fig.  103  from  Merriman  and  Jacoby's  "Roofs 
and  Bridges;"  Fig.  104  from  Johnson,  Bryan,  and 
Turneaure's  "  Theory  and  Practice  in  the  Designing  of  Mod- 
ern Framed  Structures;"  and  Fig.  155  from  Carpenter's 
"Experimental  Engineering;"  also  to  the  publishers  of 
the  Railroad  Gazette  for  Fig.  102,  and  to  the  Brown  Hoist- 
ing Machinery  Company  for  numerous  illustrations  of 
various  types  of  hoisting  machinery. 


viii  PREFACE 

In  conclusion,  the  author  wishes  to  express  his  very 
great  indebtedness  to  Mr.  Dana  Pierce,  Electrical  Engineer 
for  the  Underwriters'  Laboratories,  jointly  with  whom, 
when  associates  at  Pratt  Institute,  the  original  notes  were 
written.  Much  of  the  plan  and  present  form  of  the  text 
is  due  to  his  valuable  suggestions  and  assistance. 

Grateful  thanks  are  also  extended  to  Mr.  Arthur  L. 
Williston,  Director  of  the  School  of  Science  and  Technology, 
Pratt  Institute,  for  his  encouragement  and  many  sugges- 
tions in  developing  the  course  in  mechanics;  to  my  asso- 
ciates, Dr.  Harrison  H.  Brown,  Mr.  William  H.  Timbie,  and 
Mr.  John  A.  Randall  for  helpful  criticisms  of  the  manu- 
script, and  to  Mr.  Randall  also  for  superintending  the  pre- 
paration of  the  illustrations  and  supplying  the  index. 

J.  M.  JAMESON. 
BROOKLYN,  N.  Y. 

September,  1909. 


CONTENTS 


PAGE 

PREFACE iii-v 

CHAPTER  I 
FORCE 

Definition  of  Mechanics — Action  and  Reaction — Effects,  Measure- 
ment, and  Definition  of  Force — Positive  Actions  and  Reactions 
— Passive  Reactions — Reactions  due  to  Friction,  Elasticity, 
and  Inertia — Tension  and  Compression 1-10 

CHAPTER  II 

GENERAL  IDEAS   OF   FORCE.    GRAPHICAL   REPRESENTA- 
TION OF  VELOCITY,  FORCE,   ETC. 

Motion — Translation  and  Rotation — Velocity — Moment  of  Force 
— Positive  and  Negative  Moments — Representation  of 
Velocity  or  Force  by  a  Line — Free  Body — Force  Diagrams 
for  a  Free  Body 11-24 

CHAPTER  III 

COMPOSITION  AND  RESOLUTION  OF  FORCES,  VELOCITIES, 

ETC. 

Composition  of  Velocities — Parallelogram  of  Velocities — Parallel- 
ogram of  Forces — Computation  of  Resultant — Resolution  into 
Rectangular  Components — Use  of  Squared  Paper  in  the 
Resolution  of  Forces — Composition  of  Several  Forces  Acting 

at  a  Point 25-39 

ix 


x  CONTENTS 

CHAPTER  IV 
EQUILIBRIUM:    THREE  FORCES  AT  A  POINT 

PAGE 

Definition  of  Equilibrium — Two  Force  Pieces — Forces  at  the 
Joints  of  a  Structure — Three  Forces  at  a  Point  and  in  One 
Plane — Application  of  Parallelogram  of  Forces  to  Suspended 
Bodies  and  to  Simple  Truss  Constructions 40-49 

CHAPTER  V 
PARALLEL  FORCES.     CENTER  OF  GRAVITY 

Conditions  of  Equilibrium  for  Parallel  Forces — Applications — 
Three  Parallel  Forces — Resultant  of  System  of  Parallel  Forces 
— Couples — Center  of  Gravity — Center  of  Gravity  for  Regular 
Figures — General  Equations  for  Center  of  Gravity — Center  of 
Gravity  of  Sections — Center  of  Gravity  by  Experiment 50-61 

CHAPTER  VI 

CONCURRENT  FORCES  APPLIED  AT  DIFFERENT  POINTS. 
ALGEBRAIC   CONDITIONS  FOR  EQUILIBRIUM 

General  Conditions  for  Three  Forces — Triangle  of  Forces — Polygon 
of  Forces — The  Use  of  Simple  Trigonometry — General  Condi- 
tions for  Equilibrium  of  Concurrent  Forces — Application  to 
Simple  Hinge  Structures,  Cranes,  etc. — Equilibrium  of  Forces 
not  in  the  Same  Plane — Non-concurrent  Forces — Problems 
of  the  Door,  Ladder,  Wall  Crane,  etc. — Hinge  Reactions — 
Summary  of  Conditions  for  Equilibrium 62-87 


CHAPTER  VII 
FORCES  IN  SOME  COMMON  COMMERCIAL  STRUCTURES 

Cranes — Derricks — The   Shear   Legs — Stone   Tongs — Bridge   and 
Eoof  Trusses— Water  Tank  Supports— The  Arch 88-107 


CONTENTS  xi 

CHAPTER  VIII 
MOTION 

PAGE 

Speed— Velocity— Angular  Velocity— The  Radian— Uniform  and 
Variable  Motion — Uniformly  Accelerated  Motion — Accelera- 
tion— Angular  Acceleration — Average  Velocity — Distance  in 
Accelerated  Motion — Falling  Bodies — Projectiles — Simple 
Harmonic  Motion — Phase — Sine  Curves — Pendulums — Deter- 
minations of  "g  " 108-142 

CHAPTER  IX 
FORCES  PRODUCING  MOTION 

Mass — Inertia — Weight — Relation  of  Acceleration  to  Force  Acting 
and  to  Mass  Moved — Units  of  Force  and  of  Mass — Applica- 
tions of  the  Equation  f=ma — Motion  on  an  Incline — 
"Acceleration  Diagrams" — Computation  of  Forces  for  Given 
Accelerations — Tension  in  Hoisting  Ropes — Starting  and 
Stopping  a  Body — Rotational  Inertia — Moment  of  Inertia — 
Radius  of  Gyration — Computation  of  Torque  and  Angular 
Acceleration  for  Shafts,  Fly-wheels/  etc. — Centrifugal  Force 

143-174 

CHAPTER  X 
WORK,   POWER,   ENERGY 

Definition  of  Work — Units  of  Work:  Foot-pounds,  Erg,  Joule — 
Work  Diagrams — Power — Units  of  Power — Computation  of 
Work  and  Power — Energy — Kinetic  and  Potential  Energy — 
The  Transformation  of  Energy — Energy  of  a  Rotating  Body 
— Fly-wheels — Momentum — Conservation  of  Momentum — 
Impulsive  Forces 175-198 

CHAPTER  XI 
FRICTION.    THE  GENERAL  LAWS  OF  MACHINES 

Friction — Coefficient    of    Friction — Laws    of    Friction — Dry    and 
Lubricated  Surfaces — Friction  of  Machines,  Shafting,  Belts, 
etc. — Relation  of  Friction  to  Lubrication,  Speed,  etc. — Law 
of  Conservation  of  Energy — Input,  Output,  and  Efficiency  of 


xii  CONTENTS 

PAGB 

Machines — Velocity  Ratio — Mechanical  Advantage — General 
Law  for  a  Machine — Dynamometers — The  Prony  Brake — 
Transmission  Dynamometers — The  Winch — Gears — Sprocket 
and  Belt  Machines — The  Jack  Screw — The  Block  and  Tackle 
— The  Chain  Hoist 199-235 

CHAPTER  XII 
ELASTICITY 

Elastic  Material — Stretch  of  a  Steel  Wire  under  Tension — Stress 
— Strain — Hooke's  Law — Modulus  of  Elasticity — Elastic 
Limit — Yield  Point — Ultimate  Strength — Factor  of  Safety 
— Effect  of  Temperature  and  Repeated  Strain  upon  the 
Elastic  Properties  and  Strength  of  Material — Shear  Strain — 
Modulus  of  Rigidity — Torsion — Bending — Neutral  Axis — 
Shearing  Force  at  a  Section — Bending  Moment — Bending 
Moment  and  Shear  Diagrams — General  Equation  for  Stress 
and  its  Application — Effect  of  Dimensions  and  Shape  of 
Section  on  Stiffness  for  Beams  and  Columns — Hooks 236-269 

CHAPTER  XIII 
MECHANICS  OF  FLUIDS 

Fluids  Defined — Liquids  and  Gases — Density — Specific  Gravity 
— Pressure  in  Fluids  at  Rest  —  Transmission  of  Pressure 
through  Fluids,  Pascal's  Principle — Hydraulic  and  Pneumatic 
Machinery — Pressure  of  Fluids  on  Retaining  Walls — Center 
of  Pressure — Buoyant  Force  of  Liquids  and  Gases — Archi- 
medes7 Principle — Hydrometers — Properties  of  a  Gas — 
Atmospheric  Pressure — The  Barometer — Boyle's  Law — 
Pressure  Exerted  by  Mixed  Gases,  Dalton's  Law — Vacuum 
and  Pressure  Gauges  and  the  Measurement  of  Pressure .  .  270-298 

APPENDIX 

Useful  Numbers,  Units,  and  Constants — Significant  Figures- 
Curves  and  Curve  Plotting — The  Equation  of  a  Straight 
Line — Simple  Trigonometric  Functions — Three-place  Table 
of  Sines,  Cosines,  and  Tangents 299-316 


ELEMENTARY  PRACTICAL  MECHANICS 


CHAPTER  I 

FORCE 

1.  Mechanics. — Mechanics  treats   of  forces   and   of  the 
effects  of  forces.     In  the  following  pages  we  shall  con- 
sider:   1st.    Bodies  acted  upon  by  forces  in  such  a  way 
that   they   are   compelled   to   continue   without  change  in 
their  existing  condition  of  rest  or  motion,  i.e.,  if  at  rest, 
are  compelled  to  remain  at  rest,  and  if  moving  in  a  given 
direction  or  rotating  at  a  given  rate,   are  compelled  to 
continue  moving  in  the  same  direction  or  rotating  at  the 
same  speed.     2d.   Bodies  compelled  by  forces  to  continually 
change  their  motion,  i.e.,  made  to  move  faster  or  slower 
or  in  a  different  direction.     3d.    The  effects  of  forces  in 
changing  the  form  of  bodies  by  stretching,  compressing, 
or  bending  them. 

It  is  necessary,  therefore,  in  beginning  the  study  of 
mechanics,  to  have  a  clear  conception  of  the  meaning  of 
the  term  force.  The  following  discussion  and  illustrations 
are  intended  to  make  clear  the  significance  of  the  term  as 
used  in  practical  mechanics: 

2.  Action  and  Reaction. — Suppose  we  have  an  ordinary 
spring  balance  lying  upon   a  perfectly  smooth  horizontal 


2  ELEMENTARY   PRACTICAL   MECHANICS 

table.  By  "perfectly  smooth "  table  we  are  to  under- 
stand one  along  which  the  balance  can  be  drawn  without 
friction  (a  condition  which  of  course  can  only  be  realized 
approximately  for  any  actual  table).  If  now  we  take 
hold  of  the  ring,  we  may  slide  the  balance  steadily  along 
the  table,  once  started,  with  no  sensation  of  effort;  we 
will  be  unable  to  pull  upon  it  unless  we  increase  its  speed. 
But  suppose  we  pass  the  hook  of  the  balance  over  a  pin 
and  then  pull.  We  may  now  apply  a  force  with  the  hand, 
the  spring  will  be  extended  and  the  reading  of  the  grad- 
uated balance  will  serve  as  a  measure  of  this  action  tend- 
ing to  move  the  balance.  The  pin  therefore  provides  a 
condition  necessary  before  force  can  be  applied,  which,  if 
we  replace  the  pin  by  a  second  balance  like  the  first,  we 
find  to  consist  of  an  equal  action  in  the  opposite  direction. 
This  opposing  action  is  commonly  spoken  of  as  the  reaction. 

As  other  illustrations  of  the  same  idea,  consider  the 
following: 

Example  L — Suppose  a  rope  with  a  boy  pulling  at  each 
end.  Here  we  have  two  actions,  equal  in  amount  and 
opposite  in  direction.  One  boy  cannot  pull  unless  the  other 
provides  the  opposing  reaction. 

Example  2.  — Suppose  a  rope  tied  to  a  post  with  a  boy 
pulling  at  its  other  end.  Here  the  post  is  taking  the  place 
of  the  second  boy  of  Example  1,  and  is  exerting  the  reac- 
tion equal  and  opposite  to  the  pull  which  the  boy  is 
exerting. 

J  Example  3. — Suppose  a  book  resting  upon  a  table.     The 
earth   is   exerting   a   downward   pull    (gravity)    upon   the 
./  book,   causing  it  to  push  against  the  table,   and  at    the 
> •/  £  same  time  the  table  is  exerting  an  upward  pressure  against 
Af  /'the  book  which  is  equal  and  opposite  to  the  force  of  gravity ; 
if  the  table  failed  to  exert  this  upward  pressure  the  book 
would  fall. 

_! nirUf^  -^          ^ 

**~  -" 


ELEMENTARY    PRACTICAL    MECHANICS  3 

Example  4- — Suppose  a  horse  pulling  a  sled  at  a  uniform 
rate  over  level  ground.  Friction  in  this  case  is  exerting 
an  opposite  and  simultaneous  reaction  equal  to  the  pull 
of  the  horse. 

Example  5. — Suppose  the  horse  pulling  a  sled,  as  in 
Example  4,  increases  his  speed.  Two  reactions  now  oppose 
the  pull  applied  to  the  sled.  One,  friction,  opposes  the 
slipping  of  the  sled  over  the  ground;  the  other,  due  to 
inertia,  opposes  increase  of  speed.  These  two  together  are 
equal  and  opposite  to  the  pull  exerted  on  the  sled. 

Many  other  illustrations  might  be  given.  In  all,  we 
should  find  the  same  fundamental  conditions  present,  viz.: 

1.  Two  bodies  are  always  involved.     What  we  term  force 
is  the  "push,  pull,  attraction,  repulsion,  rubbing,  pressure, 
etc.,  of  one  body  on  another." 

2.  When  one  body,  A,  exerts  force  upon  a  second  body, 
B,   body   B  exerts   an  equal  and  opposite   action   on   A, 
called   the   reaction   of  B.      In   speaking   of   the   body   A 
alone,  we  shall  have  frequent  occasion  later  to  include  as 
among   the   "forces"   applied   to   it,   the    reaction   of  the 
second  body  B. 

3.  The  Effects  of  Force. — A  force    (i.e.,   the  pressure, 
attraction,  rubbing,  etc.,  of  some  other  body)  applied  to  a 
given  body  may  produce   one  or  both  of  two   effects  upon 
that  body: 

First,  it  may  change  the  state  of  motion  of  the  body  acted 
on;  that  is,  may  cause  the  body  to  move  if  at  rest,  or 
to  move  faster  or  slower  or  in  a  different  direction  if 
already  in  motion.  The  state  of  motion  of  a  body  under 
the  action  of  no  force  (or  of  balanced  forces)  is  either  abso- 
lute rest  or  uniform  motion. 

Or,  second,  the  force  may  alter  the  shape  of  the  body. 
For  example,  the  force  may  extend  or  compress  a  spring; 
it  may  deflect  a  beam;  it  may  compress  a  column,  as  a 


4  ELEMENTARY    PRACTICAL    MECHANICS 

weight  resting  on  a  table  compresses  the  legs  of  the  table. 
As  all  known  substances  are  compressible,  and  also  may 
be  elongated,  some  change  of  shape  is  always  produced 
by  an  applied  force,  though  perhaps  the  change  may  be 
so  slight  as  to  be  negligible. 

4.  The  Measurement  of  Forces. — The  quantity  of  either 
of  these  effects  on  any  given  body  is  found  to  depend 
directly  upon  the  quantity  of  force  applied.  Hence  a  force 
rnay  be  measured  either: 

(a)  By  the  velocity  that  it  will  impart  to  a  given  body  in 
a  given  time,  or 

(6)  By  the  amount  of  change  of  shape  which  it  produces 
in  a  given  body. 

Method  (a)  enables  us  to  find  the  force  required  to  start 
a  body  from  rest,  to  stop  a  moving  body,  or  to  .change 
the  velocity  of  a  body  by  a  stated  amount  in  a  given 
length  of  time.  (See  Chapter  IX.) 

Method  (6)  is  used  in  comparing  two  forces  by  deter- 
mining the  amount  that  each  will  stretch  a  spring,  the 
angle  through  which  each  will  twist  a  given  rod  or  wire, 
etc.  The  spring  balance  is  a  familiar  instrument  for 
measuring  forces  in  this  manner.  The  elongations  of  the 
spring  under  the  action  of  known  forces,  are  marked  on  the 
scale  of  the  instrument. 

6.  Definition  of  Force. — From  the  preceding  discussion, 
we  see  that  force  is  the  name  given  to  the  action  which  one 
body  may  exert  upon  another,  that  the  effects  of  such  actions 
may  be  either  change  of  motion  or  change  of  shape,  and, 
finally,  that  actions  between  bodies  must  be  mutual  or  in 
pairs,  hence  we  may  now  state  as  our  definition  of  force: 

11 A  force  is  one  of  a  pair  of  equal,  opposite,  and  simul- 
taneous actions  between  two  bodies  by  which  the  state  of  their 
motion  is  altered  or  a  change  in  the  form  of  the  bodies  them- 
selves is  effected." 


ELEMENTARY    PRACTICAL  MECHANICS  5 

6.  Positive   Actions   and    Reactions. — In    many    cases, 
either   one   of   the    mutual   actions   exerted   between   two 
bodies  may  be  regarded  as  the  positive  action  or  the  force , 
and  the  other  one  regarded  as  the  reaction  resisting  the 
force.     For  instance,  in  Example  1,  either  one  of  the  two 
boys  may  be  regarded  as  exerting  the  force,  and  the  other 
one  as  exerting  the  equal  and  opposite  action  spoken  of 
in  the  definition,  which  in  the  subject  of  mechanics  we  are 
calling  the  reaction. 

7.  Passive   Reactions. — There    are,   however,    three    in- 
stances in  which  we  cannot  assume  either  action  to  be 
the  force.     These  are: 

First,  Friction.  In  Example  4,  given  above,  it  is  im- 
possible for  us  to  suppose  that  the  friction  is  exerting  / 
the  force  and  that  the  horse  is  exerting  the  reaction.  Iti 
such  cases  where  friction  is  one  of  the  equal,  opposite,  anA 
simultaneous  actions,  it  is  necessary  to  regard  the  friction 
as  the  reaction  and  the  other  action  as  the  positive  action 
or  force. 

Second.  Where  the  resistance  of  some  body  due  to  its 
elasticity  or  resistance  to  a  change  of  shape  is  one  of 
the  actions,  we  cannot  regard  this  as  the  positive  action 
(unless  it  is  tending  to  produce  motion,  as  in  the  case  of 
a  spring  under  tension).  In  Example  3,  for  instance, \it/ 
would  not  be  proper  to  assume  that  the  table  was  exerV 
ing  the  force  and  that  gravity  was  exerting  the  reaction/  \ 

Third.  Where  the  resistance  of  the  body  to  change  of 
velocity  due  to  its  own  inertia  *  is  one  of  the  actions,  we 

*  INERTIA. — Inertia  is  to  be  regarded,  not  as  a  force,  but  as 
a  property  of  all  matter.  "A  body  at  rest  will  remain  at  rest, 
a  body  in  motion  will  continue  to  move  in  the  same  direction 
and  at  the  same  speed,  unless  compelled  to  change  its  state  of 
rest  or  motion  by  the  application  of  an  external  force."  The 
property  of  a  body  because  of  which  force  is  necessary  in  order 


6  ELEMENTAHY    PRACTICAL   MECHANICS 

must  usually  regard  this  as  the  reaction  (except  in  certain 
instances  where  the  velocity  of  a  body  is  being  destroyed, 
as  in  cases  of  impact  between  bodies). 

These  three  types  of  reaction  can  be  developed  by  the  appli- 
cation of  a  positive  action  or  force  only.  We  could  not 
logically  think  of  their  acting  independently  to  develop 
an  accompanying  equal  reaction.  Nevertheless,  where  the 
positive  action  does  exist,  as,  for  illustration,  in  the  case 
of  gravity  pull  upon  the  book,  Example  3,  we  may,  when 
convenient,  think  of  the  reaction  of  the  table  as  an  equal 
force  upon  the  book,  and  may  thus  speak  of  the  book  as 
under  the  action  of  two  forces,  without  regard  to  their 
source:  a  downward  force  equal  to  the  weight  of  the  book, 
and  an  equal  upward  force,  the  reaction  of  the  table.  This 
idea  of  considering  a  given  body  apart  from  its  surround- 
ings and  acted  on  by  forces  only,  while  purely  imaginary, 
is  important  in  the  solution  of  problems  in  Mechanics. 
(See  also  Art.  16.) 

8.  Tension  and  Compression. — Attention  has  already 
been  called  to  the  fact  that  actions  exist  in  pairs,  and  that 
we  have  force  only  as  one  of  the  mutual  actions  between 
two  bodies;  the  student  must  not  confuse  this  idea  and 
interpret  it  as  meaning  that  the  force  present  is  twice  its 
proper  value.  It  is  one  of  the  actions. 

Where  the  actions  are  such  as  to  tend  to  pull  the  parts 
of  the  body  apart,  the  body  is  said  to  be  under  tension.  This 
is  the  case,  for  example,  of  a  rope  tied  at  one  end  to  a  rigid 
support  and  sustaining  a  weight.  Here  gravity  is  exerting 

to  produce  change  in  its  state  of  rest  or  motion,  is  spoken  of  as 
the  inertia  of  the  body.  Whenever  the  velocity  of  a  body  is 
changing — increasing,  decreasing,  or  changing  direction — there- 
fore, we  must  consider  as  one  of  the  reactions  present,  the  resist- 
ance to  change  of  motion  due  to  the  body's  inertia.  (For  fur- 
ther discussion  of  inertia  reactions  and  force,  see  Chapter  IX.) 


ELEMENTARY    PRACTICAL    MECHANICS. 


a  downward  pull  on  the  rope  equal  to  W  and  the  support 
is  exerting  an  equal  upward  reaction,  B.  These  external 
actions  set  up  internal  actions  or  stress 
in  the  rope  and,  at  any  section  of  the 
rope,  the  actions  are  opposite  and  di- 
rected away  from  the  section,  as  a,  6, 
Fig.  1.  If,  however,  TF  =  20  Ibs.,  each 
action  is  20  Ibs.  and  we  say,  the  force 
of  the  tension  is  20  Ibs.,  not  40. 

This  is  true,  no  matter  what  the  sources 
of  the  action  and  reaction.  Thus  in  the 
structure  of  Fig.  2,  a  jointed  frame, 
ABC,  pinned  to  turn  freely  at  B,  the 
weight  at  W  tends  to  flatten  the  frame  and  cause  ends  A 
and  C  to  separate.  If  slipping  is  prevented  by  a  tie,  AC, 


FIG.  2. — Laboratory  Model  Roof  Truss. 


8 


ELEMENTARY    PRACTICAL    MECHANICS 


FIG.  3. 


and  if  C  tends  to  move  toward  the  right,  with  a  force,  P, 
of  say  10  Ibs.,  there  must  be  an  action,  F',  to  the  left  at  A 
of  also  10  Ibs.  Either  may  be  taken  as  the  action,  the 

other  as  the  reaction.  The  tie, 
AC,  is  therefore  under  a  tension 
of  10  Ibs.  (not  20  Ibs.),  as  will  be 
shown  by  the  spring  balance  at 
D. 

When  the  actions  are  such  as 
to  tend  to  push  the  parts  of  a 
body  together,  the  body  is  said  to 
be  under  compression.  An  illus- 
tration of  this  would  be  the 
column  A B,  Fig.  3,  rigidly  footed 
at  A,  and  supporting  a  load  L. 
At  any  section  of  the  column,  as 
mn,  the  actions  are  opposite  and 
directed  toward  the  section.  If  L  =  100  Ibs.,  each  action  is 
100  Ibs.,  and  the  force  of  compression  is  100  Ibs. 

9.  Action  and  Reaction  at  any  Point  in  a  Structure. — 
Equal  and  opposite  actions,  or  action  and  reaction,  are 
present  at  any  point  in  a  structure  that  we  may  select 
for  consideration.  Thus  in  Fig.  3,  action  and  reaction  are 
equal  at  B:  the  action  the  pressure  of  the  load  L,  and  the 
equal  reaction  that  due  to  the  elasticity  of  the  column 
which  causes  it  to  resist  compression  and  push  back  equally 
against  L.  Action  and  reaction  are  equal  at  the  section 
mn  or  any  other  section  across  the  column.  And  action 
and  reaction  are  also  equal  at  A:  the  push  of  the  foot  of 
the  column  downward  against  the  floor  and  the  equal 
upward  reaction  of  the  floor,  etc. 

In  the  structure  of  Fig.  2,  attention  has  already  been 
called  to  the  horizontal  actions  P  and  F',  and  the  equal 
reactions  supplied  by  the  tension  in  the  cord.  If,  ho\v- 


ELEMENTARY    PRACTICAL    MECHANICS  9 

ever,  we  consider  the  vertical  pressures   V  and   V  at   C 
and  A,  we  have  the  equal  vertical  reactions  R  and  R'. 

Or,  if  we  consider  the  pin  connecting  the  members  CB 
and  AB  at  B}  we  have  P'  and  Af  as  the  pressure  of  the 
pin  on  BC  and  AB  respectively,  and  S 
and  S'  the  equal  reactions  of  the  timbers 
on  the  pin.  S  and  S'  are  obviously  both 
holding  the  pin  up,  and  pushing  against 
one  another,  or  we  have  their  combined 
effect  as  a  vertical  reaction  M  (Fig.  4), 
equal  and  opposite  to  load  L,  thus  sup- 
porting the  pin,  and  two  equal  and  op- 
posite actions  N  and  N',  either  of  which 
may  be  regarded  as  the  action,  the  other  FIG.  4. 

as  the  reaction. 

Many  illustrations  of  the  same  character  might  be  given. 
The  first  thing  the  student  should  do  in  the  solution  of 
any  problem  is  to  find  the  actions  and  reactions  upon  the 
body  under  consideration.  These  once  properly  arranged, 
subsequent  operations  become  comparatively  simple  and 
mechanical  computations. 

PROBLEMS 

Point  out  the  actions  and  reactions  at  any  selected  point 
in  the  following: 

1.  A  weight  suspended  from  a  hook  by  means  of  a  cord. 

2.  A  man  lifting  a  heavy  weight. 

3.  A  coiled  spring  compressed  and  held  in  place  by  a 
detent. 

4.  A  car  being  drawn  at  a  uniform  rate  along  a  level 
track. 

5.  The  car  of  Prob.  4,  if  drawn  at  a  constantly  increasing 
speed. 


10  ELEMENTARY    PRACTICAL    MECHANICS 

6.  A  block  sliding  down  an  inclined  plank. 

7.  A  barge  being  towed  by  a  tug  at  uniform  speed. 

8.  A  car  being  drawn  at  uniform  speed  up  an  incline. 

9.  The  car  of  Prob.  8,  if  its  speed  is  increasing. 

10.  A  moving  ferry  boat  after  the  engines  are  suddenly 
stopped. 

11.  A  mine  car  being  lowered  uniformly  down  a  vertical 
shaft. 

12.  The  car  of  Prob.   11,  when  lowered  with  increasing 
speed. 

13.  The  lever  AC  of  Fig.  12. 

14.  The  simple  truss,  Fig.  41. 

15.  The  roller  and  plane,  Fig.  42. 

16.  The  shears,  Fig.  91. 

17.  The  arch,  Fig.  113. 

18.  The  crane,  Fig.  87. 

19.  The  system  of  pulleys,  Fig.  136,  if  A  is  moving  down- 
ward at  uniform  speed. 


CHAPTER  II 

GENERAL    IDEAS    OF    FORCE.        GRAPHICAL    REPRE- 
SENTATION OF  FORCE,  MOTION,  ETC. 

THE  mechanics  of  bodies  in  motion  will  be  treated  in 
Chapters  VIII  and  IX.  A  brief  discussion  of  terms  which 
we  shall  need  to  use  frequently  is  given  here. 

10.  Motion. — Motion  may  be  defined  as  change  of  position. 
The  change  in  position  of  a  body  is  necessarily  observed 
by  comparing  its  successive  positions  with  that  of  some 
other    body    for  the    time    being    regarded    as    fixed.     All 
motion  is,   therefore,  relative.     A  body  may  be  in  motion 
with  respect  to  one  body  and  at  rest  with  respect  to  another. 
Thus,  if  a  man  walks  down  the  aisle  of  a  car  he  is  in  motion 
with  respect  to  the  car,  but  if  he  walks  at  the  same  rate 
and  in  the  opposite  direction  to  the  motion  of  the  car, 
he  is  at  rest  with  respect  to  the  track  and  ground.     Again, 
a  body  may  possess   at  the  same  time  several  different 
motions  depending  upon  the  points  to  which  the  motion  of 
the  body  is  referred.    Thus,  suppose  a  man  is  walking  across 
a  moving  ship  which  in  turn  is  headed  straight  across  a 
stream  having  a  current.     He  has  one  motion  with  respect 
to  the  ship,  a  different  motion  with  respect  to  the  water, 
and  still  a  different  motion  with  respect  to  the  shore. 

11.  Translation  and  Rotation. — When  a  body  moves  in 
such  a  way  that  each  point  in  the  body  passes  over  a  path 
equal  and  parallel  to  the  path  of  every  other  particle,  the 
body   is   said   to   have   motion   of   pure   translation.     The 

11 


12  ELEMENTARY    PRACTICAL    MECHANICS 

motion  of  the  piston  rod  of  an  engine,  of  a  sled  over  ice, 
etc.,  are  illustrations  of  motion  of  translation. 

If,  however,  two  points  in  the  body  are  fixed  and  the 
body  rotates  about  a  line  through  these  two  called  the 
axis  of  rotation,  the  motion  is  said  to  be  one  of  pure  rota- 
tion.  A  bicycle  wheel,  raised  from  the  ground  and  set 
spinning  on  the  shaft,  the  fly-wheel  on  a  stationary  engine, 
etc.,  are  examples  of  bodies  possessing  a  pure  rotary 
motion. 

Frequently  a  body  may  have  both  motion  of  translation 
and  motion  of  rotation  at  the  same  time,  as,  for  example, 
a  wagon  wheel  when  the  wagon  is  drawn  forward,  a  car 
wheel,  etc.  general,  a  body  may  be  moved  from  one 
position  to  aAj.y  given  second  position  in  one  of  three  ways: 
(a)  By  translation  in  a  stated  direction;  (6)  by  rotation 
about  a  given  axis;  or  (c)  by  translation  a  certain  dis- 
tance in  a  given  line  and  then  rotation  through  a  definite 
angle  about  a  selected  axis. 

12.  Velocity. — By  the  velocity  of  a  body  we    mean  its 
rate   of  motion  in   any   recognized   direction.     Velocity   is 
therefore    measured   by  the  amount   of  change  of    position 
which  occurs  in  one  unit  of  time.     Thus  30  miles  per  hour 
north,  20  ft. /sec.,  etc.,  are  expressions  of  velocity. 

Since  all  motion  is  relative,  velocity  which  fixes  the 
rate  of  motion  is  also  relative,  and  a  body  may  have  at 
the  same  time  different  velocities  (either  in  amount  or 
direction)  with  respect  to  different  bodies  to  which  the 
motion  is  referred. 

13.  Moment   of  Force. — One  of  the  effects  which  may 
be  produced  by  the  application  of  force  to  a  body  has  been 
stated  to  be  change  in  the  existing  motion  of  such  body 
(Art.  3).     This  change  may  be  change  in  motion  of  trans- 
lation or  change  in  the  rotation  of  the  body,   depending 
upon  the  conditions  and  upon  the  direction  of  the  force 


ELEMENTARY    PRACTICAL    MECHANICS  13 

and  the  point  at  which  it  is  applied.  Thus  suppose  we 
pull  upon  the  rope  attached  to  a  sled;  we  may  cause  the 
sled  to  move  forward  in  the  direction  of  the  pull,  the 
rate  of  change  in  its  motion  depending  upon  the  amount 
of  the  applied  pull. 

A  force  applied  to  a  shaft  or  pulley  wheel,  etc.,  may  or 
may  not  tend  to  produce  rotation.  Thus  consider  the  drive 
wheel  in  Fig.  5.  When  the  connecting-rod  is  in  the  posi- 
tion PR,  the  force  along  it  would  tend  merely  to  push 
the  pin  P  against  the 
axle  X,  since  the  force 
line  MN  of  the  connect- 


"  ,~          u   -tM ^-— <g  4:— 44-^ gH* 

ing-rod    passes    through     y         x     yy  i — (  f v|   NX   >-£       I 

the  axle.  Any  amount 
of  force  exerted  by  the 
rod  under  these  condi- 
tions would  not  cause  the  wheel  to  rotate. 

Let  the  pin,  however,  move  to  the  position  PI,  where 
the  force  line  of  the  rod  no  longer  passes  through  the 
axle,  and  it  is  easy  to  see  that  a  force  exerted  along  the 
rod  would  tend  to  produce  rotation.  The  force  is  now 
said  to  have  a  moment  arm  (a),  which  is  the  perpendicular 
distance  of  its  force  line  from  the  axle  or  axis  of  rotation. 
(A  force  must  always  possess  this  arm  if  it  is  to  produce 
rotation.)  The  force  is  in  this  case  said  to  have  a  moment 
about  the  shaft  X  as  an  axis. 

Definition. — The  tendency  of  a  force  to  produce  rotation 
about  a  given  axis  is  called  the  moment  of  the  force  with 
respect  to  that  axis. 

14.  Measure  and  Direction  of  Moment  of  a  Force. — 
The  amount  and  direction  of  the  moment  of  a  force  depends 
upon  the  direction  of  the  force  and  upon  its  distance  from 
the  axis.  Thus  suppose  we  have  an  apparatus  such  as  shown 
in  Fig.  6,  in  which  OQ  is  a  stiff  rod  balanced  about  a  pin  P, 


14 


ELEMENTARY    PRACTICAL    MECHANICS 


and  free  to  rotate  in  either  direction  about  the  pin.     If 

now  we  hang  a  weight  W,  of  10  Ibs.  at  M,  5  inches  from 

the  pin,  this  force  will  turn  the  rod  in  a  clockwise  direc- 

.  tion  about  the  pin.     To 

L^.__ fj  slj'   *} ^I^-__ X— \\y f-- .J 

Jo IN  |M  |o 


FIG.  6. 


prevent  this  we  must 
have  an  equal  moment 
counter-clock  wise ,  and  if 
this  is  furnished  by  a 
pull  applied  by  a  spring 
balance  3,  also  attached 
5  inches  from  P,  we  shall 
find  this  balance  will  read 
10  Ibs.  (the  same  as  TF). 


If  W  is  now  moved  to  0,  10  inches  or  twice  as  far  from  Py 
we  shall  find  that  the  pull  S'  applied  at  N  must  be  increased 
to  20  Ibs.  or  twice  as  much  force.  And  if  we  wish  to  balance 
W  acting  at  0  by  an  equal  force,  we  must  apply  such  force 
at  Qj  10  inches  or  an  equal  distance  from  P. 

In  other  words,  we  can  balance  a  force  by  an  equal 
force  in  an  apparatus  like  this,  if  the  two  are  applied  at 
points  equally  distant  from  the  pin;  but  twice  the  force  is 
required  to  balance  another  force  applied  twice  as  far  from 
the  axis. 

Two  moments  are  thus  equal  when  the  product  of  force 
X  perpendicular  distance  from  axis  to  force  is  the  same  for 
both.  Thus,  by  referring  to  Fig.  6,  we  have  equal  and 
oppositely  directed  moments  under  the  following  con- 
ditions: 

(1)     WXPM=*SXPN, 
or  10X5-10X5. 

WXPO==S'XPN, 
10X10  =  20X5. 


or 


or 


(2) 
(3) 


10X10  =  10X10. 


ELEMENTARY   PRACTICAL   MECHANICS  15 

The  measure  of  the  moment  of  a  force  is  therefore  the 
product  of  the  force  by  the  perpendicular  distance  from 
the  axis  to  the  action  line  of  the  force.  Therefore, 

Moment  of  force  —  force  X  moment  arm. 

Moment   of   force   is    also   called   torque   when   applied   to 
wheels  and  pulleys. 

The  moment  arm  must  always  be  measured  along  a 
direction  perpendicular  to  the  direction  of  the  force.  In 
Fig.  6  this  is  obviously  along  the  stick.  In  Fig.  5,  how- 
ever, the  moment  arm  of  the  force  F  exerted  by  the  con- 
necting-rod is  the  distance  a  where  a  is  perpendicular  to 
the  direction  of  the  force.  F  here  both  tends  to  turn  the 
wheel  and  to  pull  it  against  the  pin.  Its  rotation  effect 
alone  is  FXa,  not  Fx distance  P'X. 

PROBLEMS 

1.  A  railroad  train  passes  a  telegraph   pole   beside  the 
track  at  rate  of  45  miles  per  hour.     What  is: 

(a)  Velocity  of  train  with  respect  to  pole? 
(6)   Of  pole  with  respect  to  train? 
How  do  these  velocities  compare  in  direction? 

2.  Two  trains  going  at  the  rate  of  15  miles  per  hour 
and  25  miles  per  hour  respectively,  in  the  same  direction, 
pass  each  other  at  a  station.     What  is: 

(a)  Velocity  of  each  with  respect  to  the  station? 
(6)  Velocity  of  second  with  respect  to  first? 
(c)   Velocity  of  first  with  respect  to  second? 
What  is  the  direction  of  the  motion  in   (b)   compared 
with  that  of  (c)? 

3.  Suppose  trains  in  Prob.  2  to  be  going  in  opposite  direc- 
tions.    What  is: 

(a)  Velocity  of  each  with  respect  to  station? 

(b)  Velocity  of  second  with  respect  to  first? 

(c)  Velocity  of  first  with  respect  to  second? 

4.  The   velocity   of   light   is    186,400   miles   per   second. 
How  long  does  it  take  light  to  come  from  the  sun  distant 
92,000,000  miles? 


16  ELEMENTARY    PRACTICAL   MECHANICS 

5.  If  the  velocity  of   sound  is  1100  ft.  a   second,  how 
far  distant  is  a  gun  whose  report  is  heard  3^  seconds  after 
the  flash  is  seen? 

6.  Express  a  velocity  of  10  miles  per  hour  in  ft. /sec. 

7.  Which  is  the  greater  velocity,  40  miles  per  hour,  or 
1800  ft. /sec.,  and  how  much? 

8.  Through  what  distance  per  minute  does  a  point  on 

the  rim  of  a  fly-wheel  move  when 

j  xp2         the  wheel  makes  150  revolutions  per 

A  j^__.    .__10-—  minute?     Diameter  of  wheel  5  ft. 

9.  AB  (Fig.  7)    is  a  straight  bar 
pinned   at   A.      Force    ^  =  50  Ibs., 
,.,      7  force  F2  =  70  Ibs.     What  is  moment 

of  each  about  the  pin? 

10.  Board  ABCD  (Fig.  8)  is  free  to  rotate  about  a  pin 
at   0.     Point  out  the  mo- 
ment arms  of  each  of  the 

indicated   forces  with   res- 
pect to  0. 

11.  Suppose     AB     (Fig. 
81)  is  a  straight  bar  weigh- 


ing 200  Ibs.,  hinged  to  wall          

at  A.     If  its  weight  acts  at     A/  V 

the  center  of  the  bar,  what     ^  \ 

is    moment  of    its  weight  Fs 

about  hinge?  If  bar  is  held  FIG.  8. 

from  rotating  by  means  of 

the  rod   CB  what  must  be  moment  of  pull  in  rod  about 

hinge?     What  is  the  tension  in  the  rod?    AB  =  6  ft.,  and 


15.  Use  of  Lines  in  Representing  Velocities,  Forces, 
etc. — A  force  is  completely  specified  by  stating:  (a)  Its 
amount;  (6)  its  direction;  (c)  the  point  at  which  it  is 
applied  to  the  given  body.  Similarly,  a  motion  is  com- 
pletely described  by  stating  its  amount,  its  direction,  and 


ELEMENTARY    PRACTICAL   MECHANICS 


17 


the  original  position  of  the  moving  body.  Since  a  line 
may  express  these  three  attributes,  i.e.,  may  have  a  definite 
length,  a  definite  direction,  and  may  start  from  or  end  at 
a  given  point,  a  line  may  be  used  to  represent  a  force,  a 
motion,  or  any  similar  directed  quantity.  In  represent- 
ing a  force  by  means  of  a  line,  the  line  is  drawn  to  or  from 
the  point  on  the  given  body  at  which  the  specified  force 
acts,  in  a  direction  corresponding  to  the  direction  of  the 
force,  and  of  a  length  to  express  the  given  force  accord- 
ing to  some  previously  determined  arid  convenient  scale, 
as,  for  example,  1  in.  =  10  Ibs.,  £  in.  =  1  lb.,  etc.  Velocities, 
motions,  etc.,  may  be  represented  by  lines  in  a  similar 
manner. 

Thus  lines  a,  b,  and  c,  in  Fig.  9,  represent  velocities 
of  15  ft. /sec.  West,  20  ft./sec.  South,  and  30  ft. /sec.  North- 


east, respectively,  from  the  point  0,  on  the  scale  of  \  in.= 
5  ft./sec,  i.e.,  line  a  is  f  in.  long,  line  &  1  in.,  and  line  c  1J  in. 
Lines  F\,  F2,  and  F3  (Fig.  10)  represent  forces  of  6,  8, 
and  10  Ibs.  respectively,  applied  to   the   rod   A B   at   the 


18  ELEMENTARY    PRACTICAL    MECHANICS 

points     PI,     PI,    and    P3,    and    acting    in    the    directions 
indicated  scale:    \  in.  =  l  lb.  (i.e.,  FI   |  in.  long,  F2  1  in., 

F3  H  in.). 

Fig.  11  represents  graphically  the  forces  acting  upon  a 
body  weighing  100  Ibs.,  being 
drawn  at  uniform  speed  up  a 
plane  A B  inclined  30°  to  the  hori- 
zontal by  a  pull  P  parallel  to  the 
plane,  friction  being  equal  to  a 
pull  of  30  Ibs.  The  pull  required 
is  80  Ibs.  R  represents  the  reaction 
FIG.  11.  of  the  plane  in  supporting  a  part 

of  the  weight,  W,  of  the  body. 

EXERCISES 

INSTRUCTIONS  FOR  QUANTITATIVE  GRAPHIC  WORK 

In  all  graphical  work  where  actual  values  are  to  be 
shown,  observe  the  following  instructions: 

(a)  Have  the  pencils  and  instruments  in  good  working 
order.  The  same  care  in  making  drawings  will  be  exacted 
here  as  in  regular  drawing  work.  Broad  lines  and  careless 
work  cannot  be  accepted. 

(6)  Determine  scale  to  which  forces,  velocities,  etc.,  are 
to  be  drawn.  This  should  be  some  simple  ratio,  as  1, 
•J,  ^,  -J-  in.  to  a  unit.  Avoid  thirds  of  an  inch  and  other 
such  inconvenient  subdivisions.  If  metric  units  are  used, 
do  not  mix  decimals  with  common  fractions. 

(c)  State   the   scale   used.     The   drawing   should   in   all 
cases  be  as  large  as  possible  with  the  page  at  your  disposal. 
One  diagram  on  a  page. 

(d)  Arrange   the   figure   so   that   it   will   come   approxi- 
mately in  the  center  of  the  page,  leaving  a  good  margin 
for  binding  at  the  left. 


ELEMENTARY   PRACTICAL   MECHANICS  19 

(e)  Put  your  name  on  every  sheet  and  indicate  clearly 
the  number  of  the  problem,  but  do  not  write  out  the 
problem  on  your  drawing.  If  requested,  submit  the  prob- 
lem on  another  sheet. 

(/)  Place  on  lines  their  values  expressed  in  force  units, 
velocity  units,  etc.  (not  their  actual  lengths  in  inches, 
cm.,  etc.),  and  give  value  of  all  angles  that  are  furnished 
or  determined.  Show  directions  by  arrows. 

(g)  If  the  value  of  a  line  or  angle  is  to  be  determined 
by  your  drawing,  distinguish  it  clearly  in  some  way  from 
the  lines  and  angles  given  by  the  problem,  and  state  its 
value. 


1.  Represent  by  lines  starting  from  a  common  point, 
and  showing  direction  and  amount: 

(a)  Velocities  of  60  ft./sec.  South;  15  ft./sec.  West; 
30  miles  per  hour  Southeast. 

(6)  Forces  of  30,  15,  8,  and  10  Ibs.,  acting  at  a 
point,  the  angle  between  the  first  and  second  being 
30°,  that  between  second  and  third  50°,  that  between 
third  and  fourth  75°. 

2.  If    15  Ibs.    are   represented   by   a   line   2J    in.   long, 
what  will  a  line  6  in.  long  represent?     How  long  a  line 
will  be  required  to  represent  108  Ibs.? 

3.  From  a  straight  rod,  AB,  6  ft.  long,  forces  act  up 
and  down.     The  forces  up  are  as  follows:  6  Ibs.  at  distance 
1  ft.  from  A;    8  Ibs.  1J  ft.  from  A;    12  Ibs.  1  ft.  from  B. 
The  forces  down  are  as  follows:   18  Ibs.  3  ft.  from  A;  5  Ibs. 
2ft.   from  A;    10  Ibs.   at  A.     Represent  this  system  by 
diagram  drawn  to  scale,  stating  scales  used.     (Use  sepa- 
rate scales  for  force  scale  and  apparatus  scale.) 

4.  Draw  to   scale   lines  representing   force  of-  2000  Ibs. 
pulling  a  car  straight  along  a  track,  a  force  of  300  Ibs. 


20  ELEMENTARY   PRACTICAL   MECHANICS 

holding  it  back,  and  a  force  of  80  Ibs.  pushing  it  sidewise 
against  the  rails. 

5.  (a)  Show  by  diagram  to  scale  an  inclined  plane  50  ft. 
long,  rising  30°  from  the  horizontal. 

(b)  Draw  height  and  base  of  this  plane.     Determine  by 
measurement  the  length  of  each. 

(c)  Show  a  force  of  36  Ibs.  acting  at  angle  of  20°  with 
the  face  of  the  plane. 

16.  Free  Body. — As  stated  in  the  preceding  article, 
forces  are  the  actions  exerted  by  bodies  upon  one  another. 
We  can  apply  forces  to  a  body  only  by  getting  other 
bodies  to  act  upon  it. 

It  is  often  convenient  in  mechanics,  however,  to  dis- 
regard the  sources  of  the  forces  acting  and  to  conceive 
of  a  body  from  which  all  external  bodies  that  act  upon 
it  have  been  removed  and  their  places  taken  by  the  force 
which  they  exerted.  In  such  cases  it  is  convenient  to 
include  all  reactions,  whether  positive  or  passive,  among 
the  forces,  giving  them  their  proper  direction  with  respect 
to  the  body  under  consideration,  and  assigning  to  them 
a  value  equal  to  their  opposite  force  action.  A  body 
thus  regarded  as  acted  upon  by  forces  only,  is  called  a  "free 
body." 

Thus  in  the  case  of  the  lever  AC  used  in  applying  forces 

to  a  screw  jack,  Fig. 
^     12 j  we  may,  when  con- 
venient,   regard   all 
supports,  etc.,  as  re- 
moved    and     their 
places    taken    by  the 
u forces  "  with    which 
FlG-  12«  they    act    upon    AC. 

We  then  have  the  "  free  body"  and  arrangement  of  forces 
shown  in  Fig.  13.     The  elastic  reaction  of  the  head  is 


ELEMENTARY  PRACTICAL  MECHANICS  21 

replaced  by  the  force  R,  and  this  with  the  applied  load  L, 
the  pull  of  the  rod  at  A  and  the  weight  W  of  the  lever,  are 
all  the  forces  acting  upon 
the  lever.     They  are,   of  A 

course,      merely       shown   , 

J  A |B D c 

qualitatively  here.     If  the  'i  •  i' 

load  L,  dimensions  of  the 

lever,  etc.,  be  known,  as 

in  cases  to  be  considered   P  vt  •  L 

later,  all   forces    may  be  „      ^ 

determined     in     amount, 

and  the  lines  for  P,  R,  W,  and  L  may  be  drawn  to  scale, 

showing  the  amounts  of  the  forces. 

The  pulley  at  L  in  the  derrick,  Fig.  15,  is  shown  as  a 

"  free  body,"  in  Fig.   14. 

In  larger  structures  containing  several  parts,  any  par- 
ticular member  may  be  taken  as  the  "free 
body  "  as  desired. 

Thus  in  the  derrick  of  Fig.  15,  for  example, 
we  may  consider  the  jib  CD  as  a  "  free  body/' 
the  other  parts  being  replaced  by  their  forces, 
as  in  Fig.  16  (a).  T  indicates  the  force  applied 
by  the  pull  in  the  tie  DE,  L  the  force  due 
to  the  load  at  D,  R  the  reaction  at  (7,  W 
the  weight  of  the  jib,  and  P  the  tension  in 
the  hoisting  rope  running  from  D  to  the 
drum  of  the  hoisting  engine.  (This  rope  is 
not  shown  in  the  photograph.)  If  we  neglect 

friction,  P  is  evidently  equal  to  \L. 

With  a  known  load  at  L,  known  angles,  etc.,  the  amounts 

of  these  forces  may  be  found  as  we  shall  see  later. 

In  many  cases  the  exact  direction  of  the  reaction  R  may 

not  be  known.     It  is  evident,  however,  that  the  hinge  at  C 

keeps  the  end  of  the  jib  from  dropping  down  or  being  pushed 


22 


ELEMENTARY  PRACTICAL  MECHANICS 


o  15.  —  Derrick  Car. 


toward  the  right.     These  two  effects  of  the  pin  may,  in  such 
cases,  be  shown  in  place  of  R,  as  V  and  H  of  Fig.  16  (6). 


>T 


r 


FIG.  17. 

It  is  evident  from  the  preceding  that  the  jib  is  under 
compression. 


ELEMENTARY  PRACTICAL  MECHANICS  23 

If  we  regard  mast  A B  as  the  free  body,  we  have  the 
arrangement  of  forces  shown  in  Fig.  17.  Tt  H,  and  V 
are  the  forces  exerted  upon  the  mast  by  the  tie  DE  and 
jib  CD  respectively.  They  are  the  equal  and  opposite 
reactions  to  the  forces  exerted  by  the  mast  on  the  jib,  as 
in  Fig.  16.  Note  their  opposite  directions  in  the  two 
figures. 

RI  and  R2  are  the  floor  reactions  upon  the  mast,  PI 
and  P2  the  pulls  exerted  by  the  guys  AF  and  AG,  and  W 
the  weight  of  the  mast. 

The  tie  ED  regarded  as  a  free  body  has  the  forces 
C,  L,  and  T,  acting  upon  it 
as  shown  in  Fig.  18.  The 
effect  of  C  and  L  is  that  of 
a  single  force  T\  and  ED  is 
therefore  a  simple  tension 
member. 

If  desired,  the  pin  at  D  alone  may  be  taken  as  the  free 
body,  when  the  force,  diagram  of  Fig.  19  results. 

Or,  if  more  convenient,  the  crane  may  be  considered  as 
D^  a  whole  to  be  one  body  and  the  force 

r  diagram  of  Fig.  20  constructed.  The 
external  forces  are  RI,  R2,  L,  P,  PI, 
P2,  and  weights  of  parts  W  and  W\. 
The  internal  actions  at  the  pin  C  and 

i-  in  the  tie  ED  need  not  be    considered, 

for  each  consists  of  an  equal,  opposite, 
pair  which,  therefore,  just  annul  one  another,  as  far  as 
external  effect  upon  the  whole  crane  is  considered. 

The  student  should  study  these  diagrams  of  the  forces 
for  the  members  of  a  structure,  very  carefully,  and  should 
note  particularly  the  dependence  of  the  direction  of  the 
force  upon  the  particular  member  considered.  He  should 
construct  diagrams  for  various  given  structures,  until 


24  ELEMENTARY  PRACTICAL  MECHANICS 

thoroughly   familiar  with   the   graphical   representation   of 
force  conditions  at  any  selected  point.     Upon  the  thorough 
comprehension  of  these  ideas  depends 
the  success  with  which  the  solution 
of  subsequent  problems  in  mechanics 
may  be  attempted.   Where  the  weight 
of  any  part  is   included  among  the 
forces  acting  upon  the  body  it  is  to 
be  considered  as  a  single  force  act- 
ing at  a  point   called    the  center  of 
gravity  of  the  body.     The  center  of 
gravity  of  all  regular  bodies-  is  situ- 
ated at  the  center  of  figure.     In  other  cases  its  location 
will  be  stated. 

EXERCISES 
QUALITATIVE  FORCE  DIAGRAMS  FOR  A  "FREE  BODY" 

(Point  of  application  and  direction  of  forces  to  be  shown,  but 
lines  need  not  be  drawn  to  scale  to  show  value.) 

1.  Construct  force   diagrams  for  the  problems   1-13   of 
Chapter  I. 

2.  Construct  a  force  diagram  for  a  body  W  being  drawn  with 
uniform  speed  upon  an  incline  rising  40°  to  the  horizontal. 

3.  Construct  force  diagrams  for  the  following,  as  "  free 
bodies :" 

(a)  Arm  A B  of  wall  crane,  Fig.  81. 
(6)   Boom  BA  and  tie  BC  of  model  of  hoisting  crane, 
Fig.  76. 

(c)  Stick  CB  and  tie  AB  of  the  apparatus  shown  in 
Fig.  72. 

(d)  Mast  EO  of  crane,  Fig.  85. 

(e)  Boom  of  derrick,  Fig.  88. 


CHAPTER  III 

COMPOSITION  AND  RESOLUTION  OF  FORCES, 
VELOCITIES,  ETC. 

17.  Composition  of  Velocities. — In  a  preceding  para- 
graph it  was  pointed  out  that  a  body  might  have  one 
velocity  with  reference  to  a  second  body  and  the  second 
body  might  have  some  other  velocity  with  reference  to  a 
third.  In  such  cases  ii>  is  often  important  to  determine 
what  the  velocity  of  the  first  body  is  with  reference  to  the 
third.  This  velocity  can  be  determined  by  properly  com- 
bining the  first  two  velocities.  The  velocity  obtained  in 
this  way  is  called  the  Resultant  Velocity,  and  the  process 
by  which  it  is  obtained  is  called  the  Composition  of  Veloci- 
ties. The  two  velocities  from  which  the  resultant  is 
obtained  are  called  the  Component  Velocities.  The  re- 
sultant velocity  shows  the  direction  and  rate  of  motion 
of  a  body  to  which  two  velocities  different  in  amount 
or  direction  or  both  are  simultaneously  imparted. 
Two  cases  arise  in  the  composition  of  velocities: 
CASE  I. — The  velocities  are  along  the  same  straight  line. 
It  is  evident  here,  from  a  little  consideration,  that  if  the 
two  component  velocities  are  in  the  same  direction  their 
resultant  equals  their  sum.  If  the  two  velocities  are  in 
opposite  directions  their  resultant  equals  their  difference. 
Or,  to  make  a  general  statement  which  applies  to  the 
composition  of  any  number  of  velocities  in  the  same  straight 
line,  whether  in  the  same  direction  or  opposed:  The  resul- 

25 


26  ELEMENTARY  PRACTICAL  MECHANICS 

taut  is  always  equal  to  the  algebraic  sum  of  the  component 
velocities. 

Thus,  if  a  man  walk  at  the  rate  of  4  miles  per  hour  down 
the  aisle  of  a  car  going  20  miles  per  hour,  his  velocity 
with  respect  to  the  ground  will  be:  (1)  If  he  walks  in  the 
direction  the  car  is  going,  4  +  20  =  24  miles  per  hour;  (2) 
if  he  walks  in  the  opposite  direction,  20  —  4  =  16  miles  per 
hour.  Or,  calling  motion  in  direction  of  car  positive, 
motion  in  contrary  direction  negative,  we  have  for  the 
first  case  (+4)  +  (  +  20)=  +24,  the  sign  showing  the  direc- 
tion of  the  resultant  velocity;  for  the  second  case  (  —  4)  + 
(  +  20)  =+16.  A  minus  sign  before  the  resultant  would 
indicate  that  the  combined  result  gives  a  motion  in  oppo- 
site direction  to  that  of  the  car.  Thus  in  (2),  if  the  velocity 
of  the  car  were  only  2  miles  per  hour,  we  should  have  as 
our  expression  ( —  4)  +  ( +  2)  =  —  2,  or  the  man  is  traveling 
2  miles  per  hour  with  respect  to  the  ground,  and  in  a 
direction  opposite  to  that  in  which  the  car  is  going. 

CASE  II. — The  velocities  are  along  lines  making  an  angle 
with  each  other.  The  component  velocities  must  here  be 
added  geometrically  instead  of  algebraically.  This  process 
will  be  explained  in  the  following  discussion  of  the  paral- 
lelogram of  velocities. 

18.  Parallelogram  of  Velocities. — In  Fig.  21  imagine  the 
point  0  to  be  moving  with  a  velocity  Y  along  the  blade 
of  the  T-square  AB.  If  the  T-square  is  at  rest,  in  one 
unit  of  time  the  point  will  move  along  the  path  OP  to  P. 
If,  on  the  contrary,  point  0  is  at  rest  on  the  blade,  but 
the  whole  T-square  slides  along  the  drawing  board  to  a 
new  position  CD,  with  a  velocity  X,  in  one  unit  of  time 
0  will  pass  along  the  path  OR  to  R.  Finally,  if  point  0 
moves  along  the  blade  with  a  velocity  Y  and  at  the  same 
time  the  T-square  slides  along  the  drawing  board  with 
the  velocity  X,  in  one  unit  of  time  the  point  0  will  be 


ELEMENTARY  PRACTICAL  MECHANICS 


27 


carried  along  the  board  a  distance  OR  and  across  the 
board  a  distance  OP,  or  it  will  be  at  Q,  which  is  the  farther 
extremity  of  the  diagonal  of  the  parallelogram  OPQR, 
shown  dotted  in  Fig.  21. 

If  a  unit  of  time  half  as  big  had  been  chosen,  point  0 
would  have  moved  half  as  far  along  the  blade,  and  the 
T-square  would  have  carried  it  half  as  far  along  the  board. 
The  point  0,  therefore,  in  half  the  time  would  have  moved 
to  S.  If  the  unit  of 
time  is  still  further  sub- 
divided, other  points,  as 
T,  W,  etc.,  may  be  lo- 
cated. All  these  points 
will  lie  on  the  diagonal 
OQ  of  the  parallelogram. 
Hence  the  path  traveled 
by  the  point  0  to  which 
the  two  velocities  X  and 
Y  are  imparted,  is  along 
the  diagonal  of  the  paral- 
lelogram; and  since  the  position  Q  is  reached  in  unit  time, 
the  length  of  this  diagonal,  OQ,  represents  correctly  the 
resultant  velocity  of  the  point. 

Therefore,  to  find  the  resultant  velocity  of  a  body  to  which 
two  component  velocities  not  in  the  same  straight  line  are 
imparted,  represent  the  two  component  velocities  in  their 
proper  directions  and  to  a  convenient  scale,  as  the  adjacent 
sides  of  a  parallelogram.  Complete  the  parallelogram,  and 
from  the  point  of  intersection  of  the  sides  representing  the 
component  velocities,  draw  the  diagonal.  This  diagonal 
represents  the  resultant  velocity  both  in  direction  and  in 
amount,  in  accordance  with  the  scale  assumed. 

19.  Composition  of  Forces. — The  principles  just  stated 
for  velocities  apply  to  the  composition  of  any  quantities 


28  ELEMENTARY  PRACTICAL  MECHANICS 

which  may  be  represented  by  lines,  i.e.,  which  have  amount 
and  direction.  The  case  of  chief  importance  in  mechanics 
is  the  composition  of  forces. 

When  two  or  more  forces  act  at  a  single  point  *  on  a 
body,  there  is  a  single  force  which  will  produce  the  same 
effect  as  the  several  forces.  This  single  force  is  called  the 
resultant  force.  Any  one  of  the  original  forces  is  called 
a  component  force. 

If  the  component  forces  all  act  in  the  same  straight  line, 
the  resultant  force  equals  their  algebraic  sum.  Its  direction 
is  readily  determined  as  for  velocities  by  a  proper  observ- 
ance of  signs. 

//  two  forces  act  at  one  point  but  in  different  directions, 
their  resultant  may  be  found  by  application  of  the  parallelo- 
gram law. 

This  may  be  stated,  in  full,  for  forces  thus: 

Parallelogram  of  forces:    IF  TWO  FORCES   ACTING   AT  A 

SINGLE  POINT  ON  A  BODY  BE  REPRESENTED  IN  MAGNITUDE 
AND  IN  DIRECTION  BY  TWO  STRAIGHT  LINES  DRAWN  FROM 
THAT  POINT,  AND  IF  A  PARALLELOGRAM  BE  CONSTRUCTED 
ON  THESE  LINES  AS  ADJACENT  SIDES,  THE  resultant  OF  THE 
TWO  FORCES  WILL  BE  REPRESENTED  IN  BOTH  MAGNITUDE 
AND  DIRECTION  BY  THE  diagonal  OF  THIS  PARALLELOGRAM, 
DRAWN  FROM  THE  POINT. 

This  proposition  may  be  regarded  as  a  corollary  of  the 
parallelogram  of  velocities,  since  we  know  that  forces 
may  be  measured  by  the  velocities  which  they  will  impart 
in  a  given  time. 

In  Fig.  22  let  AB  and  AC  represent  two  forces  PI  and 
PI,  acting  at  th6  point  A  on  some  body;  then  according 
to  some  scale  AB  will  also  represent  the  velocity  which  the 
force  PI  would  impart  in  one  unit  of  time  to  the  body 

*  NOTE. — The  point  of  application  of  a  force  may  be  considered 
as  being  at  any  point  along  the  line  of  action  of  the  force. 


ELEMENTARY  PRACTICAL  MECHANICS  29 

if  acting  alone;  and  AC  will  also  represent  the  velocity 
which  the  force  P2  would  impart  to  the  body  in  the  same 
unit  of  time  if  it  were  to  act  alone.  The  resultant  of 
these  two  velocities,  AB  and  AC,  is  the  diagonal  of  the 
parallelogram  AR,  that  is,  the  two 
forces  acting  together  will,  impart  to 
the  body  in  one  unit  of  time  a  velocity 
equal  to  AR.  But  a  force  equal  to 
AR  would  also  impart  this  same 
velocity.  Therefore,  the  force  AR, 
the  diagonal  of  the  'parallelogram,  FIQ  29 

produces  the  same  effect  as   the  two 

original  forces,  PI  andP2,  and  consequently  is  their  resultant. 
20.  Composition  of  More  than  Two  Forces. — Any  number 
of  forces  acting  at  a  single  point  and  lying  in  the  same  plane, 
may  be  combined  by  repeated  applications  of  the  parallelo- 
gram law.  In  this  method,  the  resultant  of  two  of  the 
forces  is  first  found,  then  the  resultant  of  this  first  re- 
sultant and  the  next  force  is  found,  and  so  on,  until,  all  the 
forces  have  been  combined. 

NOTE. — Computation  of  the  Resultant  of  Two  Forces,    (a)  Forces 
acting  at  90°.      If  the   parallelogram   constructed   in   the   com- 
position of  two  forces  is  also  a  rectangle, 
half  of  the  parallelogram  forms  a  right  tri- 
angle,   as  triangle   ABC,  Fig.  23.      Since 
F!          ^^  side  BC  =  AD  =  force  F2,  it  is  obvious  that 

(b)  Forces  acting  at  any  angle.     Forces 
a  and  b  acting  at  the   acute  angle  x  are 
FIG.  23.  given;  required  to  compute  their  resultant  R. 

Prolong  side  AD  and  drop  a  perpendicu- 
lar, p,  from  C  to  this  line.  Let  distance  from  D  to  foot  of 
perpendicular  at  E  =  q.  Then  ACE  is  a  right  triangle,  and 
R*  =  (b  +  q)  2 + p\  Therefore, 


30 


ELEMENTARY  PRACTICAL  MECHANICS 


But  q  =  acosx  (see  Appendix)    and  p2  =  a2  —  q2.     Substituting 
those  values  in  preceding  equation, 


cos 
or,  R2  =  a2  +  b2  +  2ab  cos  x. 

Hence,  the  square  of  the  resultant  of  two  forces  equals  the  sum 
of  the  squares  of  the  two  forces  plus 
twice  their  product  into  the  cosine  of  the 
angle  included  between  their  action  lines. 

This  equation,  written  here  for  an  acute 
angle,  is  perfectly  general,  as  will  be 
seen  from  the  following  cases: 

(1)  a  and  b  act  in  the  same  straight 
line  and  in  same  direction.  The  in- 
cluded angle,  x,  is  now  zero,  and  since 


FIG.  24. 


cos  0  =  1,  our  equation  becomes 


or, 


R  =  a  +  b. 


(2)  a  and  b  act  in  a  straight  line,  but  in  opposite  directions. 
x  now  =180°,  and  since  cos  180°=  —1,  we  have 


or, 


=  a-b. 


FIG.  26. 
(3)  a  and  b  act  at  90°.     Cos  x  now  equals  zero,  and 


as  for  a  right  triangle. 

(4)  a  and  b  act  at  an  obtuse  angle.  Cos  x  is  now  negative  with 
a  value  somewhere  between  0  and  —  1.  Therefore  our  equation 
becomes 


ELEMENTARY  PRACTICAL  MECHANICS  31 

PROBLEMS  IN  COMPOSITION  OF  VELOCITIES  AND  FORCES 

SOLVE   GRAPHICALLY 

1.  A  man  is  rowing  with  a  velocity  of  6  miles  an  hour 
at  right  angles  to  a  current  which  has  a  velocity  of  2^ 
miles  an  hour.     Find  his  velocity  with  reference  to  the 
shore. 

2.  Find  graphically  the  resultant  of  the  following  pairs 
of  forces: 

(a)  P1==36  Ibs.  and  P2  =  24  Ibs.;  angle  between  =  30°. 
(J)  P1==45  Ibs.  and  P2  =  18  Ibs.;  angle  between  =  45°. 

(c)  PI  =  22  Ibs.  and  P2  =  50  Ibs.;  angle  between  =  60°. 

(d)  P1=32  Ibs.  and  P2  =  14  Ibs.;  angle  between  =  120°. 

(e)  PI  =  25  Ibs.  and  P2  =  48  Ibs.;  angle  between  =  75°. 

3.  Draw  an  angle  of  40°  and  upon  its  sides  lay  off  forces 
of  8  and  6  units  respectively.     Find  their  resultant. 

4.  Forces  of  15  and  6  grams  act  from  a  point  at  an 
angle  of  60°.      Find  the  resultant  and  the  angle  between 
resultant  and  each  force. 

5.  Resultant  is  a  force  of  25  Ibs.     Force  a  equals  12  Ibs. 
Angle  between  a  and  resultant  is  30°.     Find  force  b  and 
angle  between  a  and  b. 

6.  Two  forces  act  at  an  angle  of  60°.     Their  resultant 
is  40  Ibs.  and  one  of  the  forces  is  25  Ibs.     Find  the  other 
force. 

7.  Three  posts  are  placed  in  the  ground  so  as  to  form 
an  equilateral  triangle.     An  elastic  cord  is  stretched  around 
them,  the  tension  of  which  is  25  Ibs.     Find  the  force  press- 
ing on  each  post  and  its  direction. 

8.  The  water  is  pressing  horizontally  against  the  side 
of  a  dam  with  a  force  of  225  tons.     The  weight  of  the 
dam  is  850  tons.     Find  the  direction  and  the  magnitude 
of  the  resultant. 

9.  A  building  is  220  ft.  high,  with  an  exposed  side  110  ft. 
long.     In  a  storm  the  wind  pressure  on  the  side  of  this 


32  ELEMENTARY  PRACTICAL  MECHANICS 

building  is  50  Ibs.  per  sq.ft.  The  weight  of  the  building 
is  10,000  tons.  Find  the  direction  and  the  magnitude  of 
the  resultant. 

10.  Suppose  a  man  is  walking  at  the  rate  of  3  miles 
an  hour  on  the  deck  of  a  steamer  in  a  direction  at  right 
angles  to  its  length,  and  suppose  the  steamer  is  moving 
through  the  water  with  a  velocity  of  12  miles  per  hour 
in  an  easterly  direction,   and  suppose  there  is  a  current 
which  is   running  at  the  rate   of  4   miles  per  hour  in   a 
southwesterly    direction.     Find   the    velocity    of   the    man 
with  reference  to  the  shore. 

11.  A  train  is  moving  at  the  rate  of  35  miles  an  hour 
in    a    northerly    direction.     The    wind    is    blowing    at    the 
rate  of  22  miles  per  hour  from  the  southwest.     Find  graphi- 
cally the  direction  in  which  a  weather  vane  would  point 
if  it  were  attached  to  the  top  of  the  train. 

12.  In  Example  11,  suppose  the  wind  is  blowing  with  a. 
velocity  of  40  miles  per  hour  from  the  southwest.     How 
fast  must  the  train  move  in  order  that  the  weather  vane 
shall  point  towards  the  northwest? 

13.  Suppose  a  man  is  rowing  at  the  rate  of  4  miles  an 
hour  and  wishes  to  cross  a  stream  at  right  angles  to  the 
bank;   suppose  also  that  there  is  a  current  of  2^  miles  per 
hour.     In  what   direction   must  he  row  with  reference  to 
the  current  in  order  that  he  may  accomplish  his  purpose? 

14.  A  boat  is  moored  in  a  stream  by  two  ropes,   one 
fastened  to  either  bank.     The  ropes  make  an  angle  of  90° 
with  each  other.     The  force  of  the  stream  on  the  boat 
is  450  Ibs.,  and  the  pull  on  one  of  the  ropes  is  200  Ibs. 
Find  the  pull  on  the  other. 

SOLVE   BY  COMPUTATION 

15.  Two  forces,  60  and  91  Ibs.,  act  at  an  angle  of  90°. 
Compute  their  resultant. 

16.  Two  forces  act  at  an  angle  of  90°.     Their  resultant 
is  200  grams  and  one  force  is  85  grams.     What  is   the 
other  force? 


ELEMENTARY  PRACTICAL  MECHANICS  33 

17.  Find  the  resultant  of  forces  of  20  and  16  Ibs.  acting: 

(a)  At  an  angle  of  30°  with  each  other. 
(6)  At  an  angle  of  70°  with  each  other, 
(c)  At  an  angle  of  120°  with  each  other. 

21.  The  Resolution  of  Forces. — Any  quantity  having 
both  amount  and  direction,  as  a  given  motion,  velocity,  or 
force,  may  be  resolved  into  two  or  more  components  which 
will  produce  the  same  effects  as  the  given  quantity.  This 
process  is  known  as  the  resolution  of  forces  (or  motions, 
etc.). 

Thus,  in  Fig.  27,  if  OR  is  a  given  force,  the  component 
forces  represented  by  the  adjacent  sides  of  any  of  the 
parallelograms  which  maybe  formed 
on  OR  as  a  diagonal,  such  as  OA 
and  OB,  or  OX  and  OY,  will  have 
OR  for  their  resultant  and  will 
therefore  be  in  every  way  equiva- 
lent to  OR,  and  may  be  at  any  F  2- 
time  substituted  for  it. 

As  an  indefinite  number  of  parallelograms  may  be  drawn 
with  OR  as  the  diagonal,  there  will  be  an  indefinite  number 
of  sets  of  components  into  which  every  force  may  be 
resolved.  It  is  evident,  therefore,  that  in  order  to  resolve 
a  force  into  definite  components,  enough  data  must  be 
given  to  determine  the  form  of  the  parallelogram.  Thus 
if  components  a  and  b  (See  Fig.  28),  are  to  make  known 
angles  with  R,  as,  say,  30°  and  40°,  draw  R  to  proper 
scale,  then  from  0  draw  lines  of  indefinite  length,  in  the 
proper  directions  for  a  and  b.  Then  from  extremity  D  of  R, 
draw  DN  and  DM  parallel  respectively  to  a  and  6.  These 
will  cut  off  lengths  from  0,  determining  values  of  com- 
ponents a  and  b. 

If  R  is  to  be  resolved  into  forces  a  and  b    and  either  a 


34 


ELEMENTARY  PRACTICAL  MECHANICS 


or  b  is  known  both  in  amount  and  direction,  points  D  and 
either  M  or  N  are  fixed;  from  which  parallelograms  may 
be  completed  and  a  and  b  both  determined. 


o         6      ^ 
FIG.  28. 


FIG.  29. 


If  components  a  and  b  are  known  in  amount  but  not  in 
direction,  first  draw  R  to  scale.  Then  from  0,  with  radius 
b,  describe  arc  op  (see  Fig.  29),  and  from  D  with  radius 
equal  to  a,  describe  arc  st.  Their  intersection  fixes  point 
N.  Locate  point  M  in  a  similar  way,  and  complete  the 
parallelogram.  Then  measure  the  angles  between  a  and 
b,  or  of  either  with  R. 

22.  Resolution  into  Rectangular  Components. — The 
most  important  case  of  the  resolution  of  forces,  velocities, 
etc.,  is  that  in  which  the  given  quantity  is  resolved  into 
two  components  which  are  at  right  angles  with  one  an- 
other. Such  components  are  called  rectangular  components. 
For  example,  we  frequently  desire  to  know  the  horizontal 
component  and  the  vertical  component  of  a  given  force. 
Suppose  OR,  Fig.  30,  to  be  the 
force.  Construct  the  parallelogram 
OXRY  so  that  OY  will  be  vertical 
and  OX  will  be  horizontal;  then 
the  horizontal  component  of  OR  will 
be  the  vertical  projection  of  OR,  or 
OX,  and  the  vertical  component  of 
OR  will  be  the  horizontal  projection  of  OR,  or  OY. 


FIG.  30. 


ELEMENTARY  PRACTICAL  MECHANICS 


35 


It  will  be  seen  from  the  figure  that  component  OX  gives 
all  the  effect  of  OR  in  the  horizontal  direction — no  more  and 
no  less — and  that  component  OFgives  the  total  vertical  effect. 

23.  Use  of  Squared  Paper  in  Resolving  a  Force  into 
Rectangular  Components. — Where  a  given  force  is  to  be 
resolved  into  rectangular  components  graphically,  the 
most  convenient  method  is  to  draw  the  force  parallelo- 


FIG.  31. 

gram  on  squared  paper;  the  components  may  then  be 
read  directly  from  the  ruled  paper.  Thus,  in  Fig.  31, 
R  is  a  force  of  120  Ibs.  acting  at  an  angle  of  40°  to  the 
horizontal;  to  find  its  vertical  and  horizontal  components. 
Suppose  the  paper  is  ruled  in  J  in.  squares:  a  convenient 
scale  is  then  J  in.  =  5  Ibs.  Drawing  a  line  OR,  40°  to  the 
horizontal,  and  3  in.  long,  to  represent  the  force,  and 
reading  from  the  paper  the  value  of  the  components,  we 
have 

Horizontal  component  Z  =  18.5  divisions  =  92.5  Ibs. 

Vertical  component       F=15.4  divisions  =  77.0  Ibs. 


36 


ELEMENTARY  PRACTICAL  MECHANICS 


This  method  makes  it  unnecessary  to  actually  measure 
OX  and  OY,  to  determine  their  length,  or  to  use  a  pro- 
tractor, etc.,  in  order  to  get  the  components  at  right 
angles.  Nor  is  it  necessary  to  draw  the  dotted  lines  show- 
ing the  components;  these  lines  are  drawn  for  Figs.  31, 
32,  and  33  merely  to  make  the  explanation  clearer. 

24.  Composition  of  Several  Forces  by  the  Method  of 
Resolution  along*  Rectangular  Axes.  —  As  suggested  in 
Art.  20,  the  resultant  of  several  forces  acting  at  a  point 

and  in  the  same  plane,  may 
be  found  by  first  determin- 
ing the  resultant  of  any  two, 
then  the  resultant  of  this  re- 
sultant and  a  third  force, 
etc.,  until  all  the  forces  have 
been  used.  A  simpler 
method,  however,  is  sug- 


m 


FIG.  32. 


gested  by  the  preceding 
article.  Thus,  suppose  four 
forces  act  at  a  point  0. 
Force  FI  of  32  Ibs.  acts  at 
an  angle  of  20°  with  the 
horizontal;  force  F%  of  20 

Ibs.  acts  at  an  angle  of  40°  with  FI;  force  F3  of  18  Ibs. 
acts  at  an  angle  of  60°  with  F2,  and  force  F±  of  40  Ibs. 
acts  at  an  angle  of  110°  with  F3.  Required  to  find  the 
amount  and  direction  of  their  resultant.  Draw  the  forces 
in  the  proper  direction  upon  squared  paper,  ruled  to,  say, 
jk  inch,  using  any  convenient  scale,  as,  for  example, 
^  inch  =  2  Ibs.,  and  determine  the  horizontal  and  vertical 
components  of  each  force  (Fig.  32). 

Since  the  components  are  equivalent  to  the  original  forces 
and  can  replace  them  in  every  way,  forces  F\,  ¥%,  F%, 
and  F4  can  be  replaced  by  eight  forces,  viz.,  two  horizontal 


ELEMENTARY  PRACTICAL  MECHANICS 


37 


Horizontal   Comp. 

Vertical  Comp. 

Force 

Direction. 

Length. 

Value. 

Direction. 

Length. 

Value. 

Ft 

Right 

15.0div. 

30.0  Ibs. 

Up 

5.5div. 

ll.Olbs. 

F2 

Right 

5.0div. 

lO.Olbs. 

Up 

8.6div. 

17.21bs. 

F, 

Left 

4.5div. 

9.01bs. 

Up 

7.8div. 

15.  Gibs. 

F< 

Left 

12.8div. 

25.  Gibs. 

Down 

15.2div. 

30.41bs. 

forces  to  the  right  from  0  of  30  Ibs.  and  10  Ibs.  respectively, 
and  two  horizontal  forces  to  the  left  of  9  Ibs.  and  25.6  Ibs. 
respectively,  three  vertical  forces  upward  of  11  Ibs.,  17.2 
Ibs.,  and  15.6  Ibs.  respectively,,  and  one  vertical  force 
downward  of  30.4  Ibs.  We  may  now  combine  the  hori- 
zontal forces  directly  to  find  their  resultant,  which  is 

+  30  Ibs. +  10  Ibs. -9  Ibs. -25.6  Ibs.  =  +5.4  Ibs., 

or,  5.4  Ibs.  to  the  right  from  0. 

Combining  the  vertical  forces  similarly,  we  have 

+  11  Ibs.  + 17.2  Ibs.  + 15.6  Ibs.  -  30.4  Ibs.  =  + 13.4  Ibs. 

or,  13.4  Ibs.  vertically  upward  from  0. 

The  original  four  forces  are  therefore  equivalent  to  two 
forces,  one  of  5.4  Ibs.  horizontally  to  the  right  from  0,  and 
one  of  13.4  Ibs.  vertically  upward  from  0. 

Constructing  a  force  parallelogram  with  these 
two  forces  as  components  (Fig.  33),  and  using 
a  larger  scale  for  convenience  (Jg  in.  =  1  lb.),  we 
find  R,  the  resultant  of  these  forces  and  there- 
fore also  of  the  four  original  forces,  FI,  F2,  F3, 
and  F±,  to  be  very  nearly  15  Ibs.,  acting  at 
an  angle  which,  measured  with  a  protractor, 
is  found  to  be  approximately  68°  with  the  horizontal. 


38  ELEMENTARY  PRACTICAL  MECHANICS 

PROBLEMS 

RESOLUTION  OF  FORCES,  MOTIONS,  ETC. 
SOLVE  GRAPHICALLY 

1.  A   force   of   75  Ibs.   has   two   components,    a   and   fr, 
making  angles  of  30°  and  45°  respectively  with  the  force. 
Find  values  of  components. 

2.  Resolve  a  force  of  100  Ibs.  into  two  components,  one 
of  which  shall  be  a  force  of  60  Ibs.  acting  at  an  angle  of 
25°  with  the  100  Ibs.  force. 

3.  Resolve  a  force  of  35  Ibs.   into  two   components   of 
20  Ibs.  and  25  Ibs.,  and  determine  the  angle  between  the 
35  Ibs.  force  and  each  component. 

4.  A  body  weighing  16  Ibs.  rests  upon  a  smooth  surface 
inclined    30°    from    horizontal.     Find    the    pressure    per- 
pendicular to   the  surface   and  the   force   parallel  to   the 
surface. 

5.  A  canal   boat  is  towed  by  a  pull  of  120  Ibs.  at  an 
angle   of   10°  with  axis  of   the  boat.     Find,  using  squared 
paper: 

(a)  Force  urging  boat   directly  ahead  in  line   of  its 

axis,  and 

(b)  Force  urging  boat  in  toward  shore. 

6.  The  rod  R  is  thrust  against  AC 
(Fig.  34)  by  a  force  of  10  Ibs.  straight 
along  R.      What  must   be  the  force 
in  direction  AC  to  prevent  slipping? 


7.  Find  the  amount   and  direction 
of  the  resultant  of  the  combinations  JTIG.  34. 

of  forces,  shown  in  Fig.  35,  using  the 
method   of   resolution   into   X   and    Y   components.     Use 


ELEMENTARY  PRACTICAL  MECHANICS 


39 


squared  paper  as  directed  in  Arts.  22  and  23,  and  tabulate 
data  neatly. 


CHAPTER  IV 
EQUILIBRIUM.    THREE  FORCES  AT  A  POINT 

25.  Definition. — When  a  body  is  acted  upon  by  forces, 
the  effect  of  which  is  the  same  as  if  no  force  acted  so  far  as 
change  in  the  existing  state  of  motion  of  the  body  is  con- 
cerned, the  forces  are  said  to  be  balanced,  or  in  equilibrium, 

and  THE   BODY  IS  SAID  TO  BE    IN  EQUILIBRIUM.       The  follow- 

ing  illustrations  will  make  clearer  the  meaning  of  this  state- 
ment: 

1.  Suppose  a  heavy  weight,  W,  rests  upon  a  horizontal 
plank,  ABj  Fig.  36.     The  forces  acting  upon  W  are  a  ver- 
|G  tical  gravity  pull,  G,  and  the 

reaction,  R,  due  to  the  elastic 
properties  of  the  plank.  As 
W  is  supported,  R  =  G;  and 
therefore  since  the  forces  are 
oppositely  directed  their  re- 


w 


|R  therefore  since  the  forces   are 


sultant  (algebraic  sum)  is  zero. 

There  is  thus  no  unbalanced  force  acting  on  W,  and  so  far  as 
motion  is  concerned,  its  condition  is  the  same  as  if  no  forces 
were  applied  to  it.  W  is  therefore  in  equilibrium.* 

2.  Or,   suppose   a  horizontal   pull,   P,   is   moving   W  at 
uniform  speed  along  AB,  Fig.  37.     The  forces  in  the  ver- 

^  *  NOTE. — The  changes  of  form  produced  in  W  and  in  plank  AB 
are  not  here  considered.  The  bodies  are  assumed  to  be  rigid. 
In  Chapter  XII  the  effects  of  force  in  producing  change  of  form 
will  be  studied. 

40 


ELEMENTARY  PRACTICAL  MECHANICS  41 

tical  plane  are  again  a  gravity  pull;  (?,  and  the  reaction,  R, 
and,  as  before, 


The  horizontal  forces  are  P  and  the  reaction  due  to 
friction,  or  Fr.  Since  P  is  just  able  to  keep  W  moving 
uniformly,  P  =  Fr  or  P-Fr  =  0.  The  resultant  of  all  the 
horizontal  forces  on  W  is  therefore  zero,  and  the  resultant 
of  all  the  vertical  forces  is  zero;  and  as  there  is  no-  hori- 
zontal or  vertical  component,  I 

there  can  be  no  resultant  force 
anywhere  in  the  plane.  All  the 
forces  on  W  are  therefore  bal- 
anced, and  thus  have  no  effect  AC 


Fr 


w 


upon  W  as  far  as  change  of 
motion  is  concerned.  Hence, 
if  W  were  at  rest  it  would  FIG  37 

remain  at  rest,  and    if    once 

set  moving,  it  will  continue  to  move  in  the  same  straight 
line  without  change  in  velocity.  W  is  therefore  in  equilib- 
rium. 

If  pull  P  exceeded  friction,  W  would  move  in  the  direc- 
tion of  P  with  constantly  increasing  velocity;  if  P  were 
less  than  Fr,  W  would  move  slower  and  slower  and  finally 
stop.  In  either  case  its  state  of  motion  would  be  changing 
and  W  would  not  be  in  equilibrium. 

3.  Example  (2)  might  be  extended  to  the  case  of  a 
body  rotating  at  a  uniform  speed,  as,  for  example,  a  shaft 
driven  by  a  belt.  When  the  driving  effect  of  the  belt 
and  the  retarding  effects  of  friction,  etc.,  are  equal,  the 
shaft  is  in  equilibrium,  and  will  continue  to  rotate  at  the 
same  speed.  When  the  driving  effect  exceeds  the  retard- 
ing, the  shaft  speeds  up;  when  the  reverse  is  true,  it  slows 
down. 


42  ELEMENTARY  PRACTICAL  MECHANICS 

Thus,  A  BODY  AT  REST,  OR  IN  UNIFORM  MOTION  IN  A 
STRAIGHT  LINE  OR  ROTATING  AT  UNIFORM  SPEED,  IS  IN 
EQUILIBRIUM. 

26.  Application    to    Engineering    Structures. — Engineer- 
ing structures,    as   the   various   types   of  trusses,   bridges, 
cranes,   etc.,   are   designed  to   support,    rigidly,   the   loads 
applied  to  them.     It  is  evident,  therefore,  that  the  com- 
ponent parts  of  such  structures  must  be  held  at  rest  or, 
as  we  may  now  say,  in  equilibrium — in  other  words,  that 
the  loads  applied  at  any  point  or  points  must  set  up  such 
tensions  and  compressions  throughout  the  structure  that 
the    forces    applied   to    each   particular    member    shall   be 
balanced,   thus   holding   that   member   in   equilibrium.     If 
we  wish,  therefore,  to  determine  the  stresses  in  the  mem- 
bers of  such  structures,  we  have  only:   First,  to  represent  a 
selected  member  or  part  as  a  "free  body  "   (see  Chapter  II, 
Section  16);  and  then,  second,  to  determine  what  values  the 
forces  applied  to  such  part  must  have  in  order  that,  acting 
in  the  fixed  directions,  they  may  balance  each  other.     Thus 
in  the  derrick  of  Fig.  15,  we  may  find  the  tension  in  the 
tie  ED  and  the  compression  in  the  jib  CD,  by  determining 
the  values  of  the  force  upon  the  jib  to  hold  it  in  equilibrium, 
i.e.,  values  of  T  and  R,  Fig.  16,  for  a  given  value  of  L  in 
order  that  the  forces  may  be  balanced.     The  solution  of  all 
such  problems,  therefore,  depends  upon  the  application  of 
the   fundamental   laws   for   the   equilibrium   of   particular 
combinations  of  forces.     These  laws  will  be  discussed  and 
their  applications  to  practical  structures  will  be  pointed 
out  in  the  succeeding  chapters. 

27.  Equilibrium  with   Two   Forces  Acting. — It  will  be 
evident  from  our  definitions  and  discussion  of  equilibrium 
that  a  body  acted  upon  by  a  single  force,  or  by  a  system  of 
unbalanced  forces    (i.e.,    a   system   whose   resultant   is   not 
zero),  can  not  be  in  equilibrium. 


ELEMENTARY  PRACTICAL  MECHANICS 


43 


G' 


FIG.  38. 


The  least  number  of  forces  which  may  produce  equilib- 
rium is  therefore  two.  It  is  obvious  also  that  a  body  acted 
upon  by  two  forces  can  be  in  equilibrium 
only  when  the  forces  are  equal  in  amount 
and  act  in  opposite  directions  in  the 
same  straight  line. 

Thus  the  suspended  body  shown  in 
Fig.  38,  if  acted  on  by  two  forces  only, 
(i.e.,  no  friction  at  pin  P),  could  be  in 
equilibrium  only  when  the  pull  of  gravity 
G  is  in  the  same  vertical  line  with  the 
reaction  R  of  the  pin.  In  any  other 
position  G  would  have  a  moment  about 
P  which  would  cause  rotation  with  in- 
creasing speed. 

The  column,  shown  in  Fig.  3,  and  the  "  tie,"  ED,  of 
the  derrick,  Fig.  15,  are  illustrations  of  typical  "  two- 
force  pieces."  Such  members  are  always  either  simple 
tension  or  compression  members,  with  the  action  line  of 
the  forces  along  the  axis  of  the  piece.  The  student  should 
note,  here,  that  when  we  are  considering  the  forces  which 
two-force  members  exert  on  the  other  parts  of  the  same  struc- 
ture, such  forces  are  always  both  toward  or  both  away  from 
the  joints  at  the  ends  of  the  member.  Thus  in  Fig.  3, 
the  column  is  pushing  upward  at  A  and  pushing  downward 
upon  the  support  at  B.  The  tie  in  Fig.  15  is  pulling  on 
the  mast  AB  in  the  direction  ED  and  also  pulling  on  the 
jib  DC  in  the  direction  DE. 

28.  Three  Forces  in  a  Plane,  Acting  at  the  Same  Point. 
— Forces  whose  action  lines  pass  through  a  common  point 
are  called  concurrent  forces. 

Let  a  and  b,  Fig.  39,  be  two  forces  whose  action  lines 
intersect  at  0.  Combining  these  by  the  parallelogram 
law,  we  find  their  resultant  R.  If  a  third  force  is  to  act 


44 


ELEMENTARY  PRACTICAL  MECHANICS 


p 


FIG.  39. 


with  a  and  b  in  such  a  way  that  the  three  shall  balance, 

or  in  other  words,  so  that  the  combined  effects  of  a,  6, 

and  c  shall  be  zero,  it  is 
evident  that  c  must  be 
equal  and  opposite  to 
resultant  R  and  that  it 
must  act  in  the  same 
straight  line  as  R.  Force 
(c),  equal  and  opposite 
to  the  resultant  R  of 
forces  (a)  and  (6),  is 

sometimes    called    the    equilibrant  of  forces    (a)    and    (&). 

Hence  three  forces  acting  at  a  point  and  in  the  same  plane 

will  produce  equilibrium  when 

THE  DIAGONAL  OF  THE 
FORCE  PARALLELOGRAM 
FORMED  ON  ANY  TWO  FORCES 
AS  SIDES,  DRAWN  FROM  THE 
POINT  OF  CONCURRENCE,  IS 
EQUAL  AND  OPPOSITE  TO 
THE  THIRD  FORCE. 


N  OTE  . — Experimental  Test 
of  Parallelogram  Laiv. — The 
truth  of  the  parallelogram 
law  may  be  easily  tested 
experimentally  as  shown  in 
Fig.  40.  A  ring  0  is  sup- 
ported by  two  cords  OA  and 
OB,  in  which  spring  balances, 
Pi  and  P2  are  inserted;  and 
from  the  ring  is  suspended  a 

weight,  W.     Hold  a  drawing     FJQ>  40._A          tus    for    test   of 
board  covered  with  paper  be-  parallelogram  law. 

hind  the  apparatus  and  indi- 
cate  center  of  ring  0  by  a  pencil  dot.    Place   dots   also   back 


ELEMENTARY  PRACTICAL  MECHANICS 


45 


of  the  centers  of  cords  OA  and  OB  and  OW.  Connect  these 
dots  by  lines  showing  the  directions  of  the  forces,  and  record 
beside  each  line  the  value  of  the  force  in  that  direction  as  deter- 
mined from  the  balances  Pl  and  P2  and  W.  The  parallelogram 
on  lines  OA  and  OB  may  now  readily  be  drawn  to  scale.  Its 
diagonal  expressed  in  force  units  should  equal  T7,  and  its  direc- 
tion be  in  the  vertical  line  OW.  Similarly,  the  diagonal  of  the 
force  parallelogram  on  OA  and  W  as  sides,  should  be  equal  and 
opposite  to  the  force  P2,  and  the  diagonal  of  the  force  parallelo- 
gram on  OB  and  W  should  be  equal  and  opposite  to  Plt 

This  condition  may  be  used  for  the  graphical  solution  of  any 
problem  in  which  three  forces  act  at  one  point,  provided  data 
for  the  construction  of  the  parallelogram  is  obtainable. 

Example  1. — Fig.  41  shows  a  laboratory  model  of  a 
simple  truss  construction  in  which  the  horizontal  member 
AB  hinged  or  pinned  at 
Aj  and  bearing  a  load  W, 
is  supported  by  a  tie  CB. 
Suppose  W=7Q  Ibs.  Re- 
quired: (a)  Tension  in  the 
tie  BC]  (V)  thrust  of  AB 
against  the  wall.  It  is 
obvious  that  the  point  B 
is  here  in  equilibrium  un-  ' 
der  the  action  of  three 
concurrent  forces,  viz. :  1st. 
Vertical  downward  pull  of 
70  Ibs.;  2d.  Pull  in  the  tie 
which  must  be  in  the 
direction  EC  (for  if  tie 
w^ere  cut  it  is  evident 


FIG.  41. — Simple  Truss. 
Laboratory  Method. 


that  B  would  move  downward);  arid  3d.  The  reaction  of 
stick  AB  which  is  along  AB  and  evidently  in  the  direction 
indicated  by  R  (for  if  AB  be  cut,  point  B  will  fall  inward 
toward  the  wall.  Reaction  R  must  be  equal  and  opposite 
to  resultant  of  forces  W  (70  Ibs.),  and  tension  in  BC. 


46  ELEMENTARY  PRACTICAL  MECHANICS 

We  are  thus  to  construct  a  parallelogram  whose  sides 
shall  have  the  directions  B W  and  EC  and  whose  diagonal 
shall  have  the  direction  BA,  the  side  BW  being  known  in 
amount. 

Lay  off  a  distance  BM  which  shall  represent  70  Ibs.,  by 
some  convenient  scale,  e.g.,  1  in.  =  10  Ibs.  From  M  draw 
a  line  parallel  to  the  tie  to  intersect  the  action  line  of  the 
force  along  the  stick,  and  from  this  point,  Nf  draw  NO 
parallel  to  BM  to  intersect  the  action  line  of  force  in  BC. 
BO  represents  tension  in  BC  to  same  scale  that  BM  = 
70  Ibs.  Measuring  this,  suppose  we  have  £0  =  9fin. 
approximately,  or  tension  in  BC  is  97.5  Ibs. 

Reaction  R  =  resultant  BN.  Suppose  BN  measures 
6^  in.  approximately,  then  reaction  R  =  65  Ibs.  Arid 
since  reaction  R  and  thrust  against  pin  at  A  are  equal 
(Why?),  thrust  at  A  =  65  Ibs. 

Example  2. — In  the  simple  "  A  "  truss,  a  model  of  which 
is  shown  in  Fig.  2,  members  AB  and  CB  are  under  com- 
pression and  are  supporting  the  load  at  B.  Three  forces 
act  on  the  pin:  the  thrusts  of  members  AB  and  CB}  and 
the  load  L.  The  directions  of  the  forces,  and  sugges- 
tions for  the  force  parallelogram  are  given  in  the  diagram 
for  Fig.  4.  The  student  should  be  able  to  complete  the 
construction  and  determine  the  compression  in  AB  and 
CB.  The  thrusts  of  these  members  at  A  and  C  may  then 
be  resolved  into  their  horizontal  and  vertical  components 
and  the  results  may  be  checked  with  the  balance  read- 
ings at  D  and  D'.  In  structures  of  this  kind,  in  which 
the  weights  of  the  members  must  be  allowed  for,  half 
the  weights  of  AB  and  CB  are  to  be  added  to  the  load 
W  at  B}  the  other  halves  regarded  as  acting  at  A  and  C. 
The  latter  should  be  added  to  the  vertical  components 
of  the  thrusts  before  comparing  with  the  balance  read- 
ing D'.  The  horizontal  base  of  this  truss  is  provided  as 


ELEMENTARY  PRACTICAL  MECHANICS  47 

a  guide  to  keep  the  plane  of  the  truss  vertical.  One  end 
of  the  truss  (C)  is  held  by  a  pin,  the  other  end  (A)  is  held 
"  free  "  by  the  balances  D  and  D'.  If  the  truss  is  sym- 
metrical, the  vertical  component  at  C  will  equal  the  corre- 
sponding component  at  A.  If  not  symmetrical,  after 
checking  for  A,  end  A  may  be  pinned  and  end  C  then 
freed  and  the  vertical  component  measured. 

Example  3. — The  forces  acting  upon  a  body  supported 
upon  an  inclined  plane  are  shown  in  the  model  of  Fig.  42. 


FIG.  42. — Inclined  Plane  and  Roller. 

The  construction  is  here  such  that  friction  of  the  roller 
on  the  plane  is  negligible,  hence  the  reaction  P  of  the 
plane  is  perpendicular  to  the  surface.  Three  forces,  the 
pull  F,  P,  and  the  weight  W  act  upon  the  roller,  holding 


48 


ELEMENTARY  PRACTICAL  MECHANICS 


it  in  equilibrium.  The  directions  of  the  three  forces  are 
all  known  for  any  given  setting  of  the  plane.  Since  W 
must  be  equal  and  opposite  to  the  resultant  of  forces  F 
and  P,  if  a  force  OR  be  drawn  to  scale  to  represent  W 
in  amount,  the  force  parallelogram  may  be  completed, 
having  OT  and  OP  as  adjacent  sides.  These  lines  may 
then  be  measured  and  the  values  of  F  and  P  determined. 
F  may  then  be  checked  with  the  balance  reading  in  the 
cord,  and  P  by  pulling  with  a  balance  in  the  direction 
OP  until  the  roller  is  just  free  from  the  plane.  This 
apparatus  represents  the  common  case  of  a  body  sup- 
ported on,  or  being  drawn  at  uniform  speed  up  an  incline. 


PROBLEMS 

THREE  FORCES  ACTING  AT  A  POINT 
SOLVE  GRAPHICALLY 

1.  Forces  of  20  and  36  Ibs.  act  at  an  angle  of  60°.     What 
third  force  acting  with  these  will  produce  equilibrium? 

2.  A  rigid,  weightless  rod  A  B,  hinged  to  a  wall  at  end  Af 
is  held  inclined  40°  to  the  vertical  by  a  cord  running  hori- 
zontally from  end  B  to  a  point  on  the  wall  vertically  above 
the  hinge.     A  weight  of  10  Ibs.  is  hung  at  end  B.     Find 
tension  in  the  cord  and  thrust  of  stick  against  the  wall 

at  the  hinge. 

3.  If    the   breaking    strength    of  the 
cord  in  problem  2  is  30  Ibs.,  what  weight 
hung  at  B  will  be  just  large  enough  to 
break  the  cord? 

4.  In  the  apparatus  of  Fig.  43. 


FIG.  43. 


If  TF=14  Ibs.,  find  the  tension  in  AB  and  thrust  in  CB. 


AC  =8  ft. 


ELEMENTARY  PRACTICAL  MECHANICS 


49 


5.  If  in  the  apparatus  shown  in  Fig.  72,  angle  CBW=  10°, 
angle  BAC  =  5Q°,  and  weight  TF  =  400  Ibs.,  find  stresses  in 
BC  and  BA,  and  the  vertical  and  horizontal  reactions  of 
the  pier  at  C  and  at  A. 

6.  In  Fig.  44  TF  =  150  Ibs.     Find  thrusts  in  BA  and  BC 
and  the  reactions  horizontally  and  vertically  at  A  and  C. 


FIG.  44. 


FIG.  45. 


7.  ABCD,  Fig.  45,  is  a  frame  of  four  sides  having  equal 
lengths  and  pivots  at  the  four  corners.      If  angle  ACB  is 
60°,  find  the  thrust  in  each  arm  when  A  and  B  are  drawn 
together  by  a  force  of  10  Ibs.,  and  find  also  the  force  sepa- 
rating C  and  D. 

8.  In  the  truss,  Fig.  2,  AB  and  BC  make  angles  of  30° 
with  the  horizontal.     The  weight  L  is  640  Ibs.     Find  the 
stress  in  AB  and  BC.     Also  find  the  tension  in  the  tie  AC. 

9.  A  tight-rope  walker  weighing  150  Ibs.  stands  in  the 
middle  of  a  tight-rope  20  ft.  long.     The  rope  is  depressed 
1  ft.     Find  tension  in  the  rope. 

10.  The  anchor  rope  of  a  balloon  makes  an  angle  of  70° 
with  the  ground.     If  the  "  lifting  force  "  of  the  balloon 
is  300  Ibs.,  find  the  tension  in  the  anchor  rope  and  the 
horizontal  force  exerted  against  the  balloon  by  the  wind. 


L____p.___^j  1 

h^H  _____  p  _________  I 

e 


CHAPTER  V 

EQUILIBRIUM:     PARALLEL  FORCES.    CENTER  OF 
GRAVITY 

29.  Conditions    of    Equilibrium    for    Parallel    Forces.  — 

Suppose  the  rod  AB,  Fig.  46,  to  be  acted  upon  by  forces 

a,    b,    c,    etc.,    which 
are     parallel.       It     is 
evident       that       the 
effect  of  these  forces 
'    may    be    any    of    the 
following:    (1)  To  dis- 
place  AB  either  up- 
•pIG  46  ward    or    downward; 

(2)    to  cause   A  B  to 

rotate   about  some  point;     or,     (3)  to  produce  both  dis- 
placement and  rotation  at  the  same  time. 

Therefore,  in  order  that  AB  may  be  in  equilibrium,  the 
tendencies  toward  displacement  and  rotation  must  each  be 

SO  BALANCED    AS   TO   HAVE    A   TOTAL   RESULTANT   OF   ZERO. 

For  convenience,  forces  acting  upward  may  be  con- 
sidered as  +  forces,  forces  downward  as  —forces;  moment 
of  force  tending  to  produce  rotation  counter-clockwise  as  a 
-\-moment,  that  tending  to  produce  rotation  clockwise  as  a 
—  moment.  Then  there  will  be  no  displacement  of  AB 
either  upward  or  downward  when  forces  upward  =  forces 
downward]  i.e.,  a  +  b  =  c  +  d  +  e]  or,  when  the  total  effect  of 
forces  is  zero,  i.e.,  a  +  b-c-d-e  =  Q, 

50 


ELEMENTARY  PRACTICAL  MECHANICS 


51 


We  may  also  determine  the  perpendicular  distance  from 
each  force  to  ANY  POINT  along  AB,  as  P,  and  thus  find  the 
moments  of  the  forces  with  respect  to  point  P.  It  is 
evident  that  forces  c,  d,  and  a  tend  to  produce  rotation 
clockwise  about  P,  forces  b  and  e  tend  to  produce  motion 
counter-clockwise.  When  these  tendencies  are  equal,  i.e., 
when 


there  will  be  no  rotation. 

The  conditions  of  equilibrium  for  parallel  forces  are 
therefore : 

1.  SUM  OF  ALL  FORCES  =  0. 

2.  SUM  OF  THE  MOMENTS  OF  ALL  FORCES  =^0. 

30.  Three  Parallel  Forces. —  Equilibrium  with  three 
parallel  forces  acting  is  a  very  common  and  important 
case  in  mechanics.  Suppose  a  b 

and  b,  Fig.  47,  to  be  two  parallel 
forces  acting  in  the  same  direc- 
tion on  a  weightless  rod,  A  B]  it 
is  required  to  find  a  third  force, 
c,  which,  acting  with  a  and  &, 
will  produce  equilibrium.  By 
condition  (1), 

i 
a  +  b  —  c  =  0.  FIG.  47. 

Since  a  and  b  act  in  same  direction,  c  must  act  in  the 
opposite  direction  and  must  equal  a  +  b.  By  condition  (2), 
taking  moments  about  P  the  point  of  application  of  c, 


Therefore, 


52 


ELEMENTARY  PRACTICAL  MECHANICS 


or,  force  c  must  act  at  such  a  point  as  to  divide  the  dis- 
tance between  a  and  b  into  parts  d\  and  d%,  which  are 
inversely  proportional  to  the  magnitudes  of  the  forces. 

If  the  forces  act  in  opposite  directions,  as  a  and  c, 
Fig.  47,  the  third  force  b  which  will  produce  equilibrium 
equals  c  — a,  and  must  be  applied  on  the  opposite  side  of  c 

force  a     d% 

(the  greater  force)  at  a  distance  such  that  •  = — . 

force  b     di 

31.  Resultant  of  Parallel  Forces.— Since  forced  (Fig.  47) 

produces  equilibrium  with  forces  a  and  6,  it  must  oe  equal  and 

opposite  to  the  combined  effects  of  a  and  6,  i.e.,  to  their  resultant. 

Therefore    the    following    statements    follow    from    the 

preceding: 

First:  The  resultant  of  two  parallel  forces  acting  in  the 
same  direction  acts  in  a  direction  parallel  to  them  both 
and  is  equal  to  their  sum. 

Second:  This  resultant  acts  at  such  a  point  as  to  divide 
the  perpendicular  distance  between  them  in  two  parts  which 

are  inversely  proportional 
to  the  two  forces. 

The  student  should 
formulate  similar  state- 
ments for  the  case  where 
a  and  b  act  in  opposite 
directions. 

32.  Apparatus  for 
Studying  the  Laws  for 


k— 22.3  cmv 


42.50  ozs. 


36.7cm.--— *\ 


1 

"]cQ_J 

1      1 

1 

I 

^6  cra.-^4  —18.5  cm--> 

J(Cley_is,c, 
—  •  f==\ 

7.5  ozs)    < 
Wt.  < 

)f  Bar 

lOozs.  12ozs. 


Parallel  Forces. — A  con- 
venient laboratory  ap- 
paratus for  testing  the 
conditions  stated  for  the 
equilibrium  of  parallel 
forces  is  shown  in  Fig.  48.  A  meter  rod  bearing  movable  steel 
knife-edges  is  suspended  "from  brass  clevis  C.  The  upward 


FIG.  48. 


ELEMENTARY  PRACTICAL   MECHANICS  53 

force  is  measured  by  a  5-lb.  spring  balance  graduated 
to  J  oz.  Downward  forces  are  applied  by  scale  pans 
and  weights  suspended  from  small  steel  knife-edges,  which 
bear  upon  the  bar  at  definite  points.  The  weights  of  the 
meter  bar,  bearings  at  (7,  and  knife-edges  must  be  included 
in  the  downward  forces.  One,  two,  or  more  upward  forces 
may  be  used  as  desired. 

Tests  should  be  made  for  algebraic  sum  of  the  forces 
and  for  the  sum  of  the  moments  about,  (1)  one  end  of 
the  bar,  (2)  point  of  application  of  an  upward  force,  (3) 
any  point  along  the  bar  where  no  force  is  applied. 

33.  Solution  of  Problems. — The  following  problems  will 
illustrate  the  application  of  the  conditions  of  equilibrium 
for  parallel  forces. 

Example  1. — Find  the  pressure  on  each  of  the  two  sup- 
ports of  the  beam  shown  in  Fig.  49.  The  force  diagram  is 
drawn  below  the  apparatus  diagram. 
The  equation  of  forces  is, 

A  +B  =  600  Ibs., 
or  A  +  £-600  =  0.  7 

600F 

For  the  equation  of  moments  it  *A  AB 

is  best  to  take  moments  about  the  5 

point  of  application  of  one  of   the      o      2       ~~| 7 
support   reactions,    since    in    that  leoo* 

case  the  moment  of  that  reaction  is  J,IQ   4g 

zero,  and  the  unknown  value  of  that 

reaction  does  not  appear  in  the  moment  equation.  Taking 
moments  about  point  of  application  of  A,  we  have 

-(3X600)+55  =  0 

Hence,  5  =  360  Ibs.     Then  from  the  first  equation, 
A+360-600  =  0; 
A  =  240  Ibs. 


1(3)     I     (ip>  (9) 
6  J 


54  ELEMENTARY  PRACTICAL   MECHANICS 

Example  2.  —  Find  amount,  direction,  and  point  of  appli- 
cation of  resultant  of  the  system  of  parallel  forces  shown 

in  Fig.  50.     Since  this  resultant 
g  is   equal    and    opposite    to    the 

|    i  force  that  will  produce  equilibri- 

(l)        (6)    I     I          da)       um  (Art.  31),  we  may  most  con- 
veniently   apply    the    laws    for 
3        equilibrium  of  parallel  forces  and 
find  this  force  E  which  may  be 
p      50  called  the  equilibrant.     A  force 

equal  to  this  force,  applied  at  the 

same    point,  but  in  the  opposite  direction,  is  the   required 
resultant.     The  equation  of  forces  gives 

6  +  4  +  ^-5-7-3  =  0. 

Therefore,  the  equilibrant  is  an  upward  force  of  5.  And 
hence  the  resultant  is  a  downward  force  of  5. 

Taking  moments  about  0,  we  have,  if  x  =  distance  of  E 
from  0: 

-(5X1)  +  (6X3)-(7X6)+5Z+(4X9)-(3X12)==0, 


x  =  5.8. 

The  resultant  is  thus  found  to  be  a  force  5  acting  down 
at  a  point  5.8  from  0. 

34.  Couples.  —  Two  equal  and  parallel  £ 

forces  acting  in  opposite  directions  are 
called  a  COUPLE.  (See  Fig.  51.)  The 
perpendicular  distance  a  between  the 
two  forces  is  called  the  arm  of  the 
couple. 

Since  F  =  Ff  it  is  evident  from  the  laws  F      51 

which  apply  to  parallel  forces  that: 

1.  A  couple  can  jrroducq  rotation  only.     The  moment  of 


ELEMENTARY  PRACTICAL  MECHANICS  55 

a  couple  is  equal  to  the  product  of  the  arm  by  one  of  the 
forces. 

2.  A  couple  cannot  be  balanced  by  a  single  force.  In  order 
to  produce  equilibrium  a  second  couple  having  an  equal 
moment  is  required. 

35.  Center  of  Gravity.— The  center   of  gravity  of  a  body 
is  the  point  at  which  we  may  assume  the  force  of  gravity  to 
always  act:    that  is,   it  is  the  point  of  application  of  the 
resultant  of  all  those  little  parallel  forces  which  gravity  exerts 
on  the  particles  of  the  body  regardless  of  the  way  in  which 
the  body  may  be  placed  or  turned. 

If  a  body  is  homogeneous  and  symmetrical  in  shape, 
the  position  of  its  center  of  gravity  will  be  evident  from 
inspection.  Thus  the  center  of  gravity  of  a  uniform  timber  is 
on  its  central  axis  at  the  middle  of  its  length,  the  center  of 
gravity  of  a  rectangle  at  the  intersection  of  its  diagonals, 
of  a  circle  at  the  intersections  of  two  diameters,  etc. 

36.  Center  of  Gravity  of  a  Triangle. — The  center  of  gravity 
of  the  triangle  ABC,  Fig.  52,  from  symmetry  will  lie  along 
the  line  BD,  bisecting  side  AC,  but 

it  will  also  lie  on  the  line  AE  which 

bisects    the    side  BC.      The    point 

of  intersection  of    BD  and  AE  is 

therefore  the  center  of  gravity  of 

the   triangle.     This   point   will   be 

found    to    be     one-third     of    the  -^JQ   52 

distance  from  the  middle  point  of 

any  side  of  the  triangle  to  the  apex  opposite  it. 

37.  Center  of  Gravity  of  any  Figure. — Given  a  card  of 
the  shape  and  dimensions  of  Fig.  53;    required  its  center 
of  gravity. 

Assume  an  axis  XXf  (here  for  convenience  one  edge  of 
card)  and  let  C  be  the  center  of  gravity  at  a  distance  YQ 
from  this  axis. 


56  . 


ELEMENTARY  PRACTICAL  MECHANICS 


If  the  card  is  of  uniform  thickness  and  material,  the 
weight  of  each  of  the  three  rectangles  of  which  it  is  com- 
posed will  be  proportional  to  their  areas,  and  the  weight 
of  the  whole  proportional  to  its  total  area.  Since  a  force 
equal  to  its  weight  acting  at  the  center  of  gravity  of  the 
card  will  support  it,  we  may  write  as  our  equation  of 
moments  about  the  XX'  axis, 

40  XF0- (24X1.5) -(4X5)- (12X8)  =0; 
TO  =  3.8  inches. 

Or,  the  center  of  gravity 
of  the  card  is  some- 
where on  a  line  parallel 
to  axis  XX'  and  3.8 
inches  from  it. 

From  the  symmetry 
of  the  card,  we  know 
that  the  center  of  gravi- 
ty will  be  located  on 
center  line  A  B.  Hence 
center  of  gravity  of 
the  card  is  at  a  point 
C  on  AB,  3.8  inches 
from  B. 

If  the  figure  were  not 
symmetrical,  we  could 
take  moments  about  an  axis  YY'  at  right  angles  to  XX', 
and  determine  the  distance  X0  from  this  axis  to  the  center 
of  gravity  C. 
Thus  for  Fig.  53, 

40X0- (24X4) -(4X4)- (12X4)  =0; 
X0=4  inches  from  YY\ 


x — 


I 

A 

•f- 

- 

* 

+ 

\ 
* 

«- 

s.- 

_x  

-> 

"> 

C 

+ 

1" 

\k 

'0 

T 

l/r 

i 

i 

/ 

B 

-c 

" 

i     ^ 

O 

Y' 


FIG.  53. 


ELEMENTARY  PRACTICAL  MECHANICS 


57 


Or,  center  of  gravity  C  is  at  a  point  3.8  inches  from  XX' 
and  4  inches  from  YY'. 

From  the  preceding  it  is  evident  that  the  center  of 
gravity  of  any  figure  may  be  found  from  the  general 
equations : 

(1)     AX0  =  sum  (ax), 


(2)     AF0  =  sum  (ay). 

Where  A  is  the  total  area  of  the  figure,  XQ  the  distance 
from  center  of  gravity  of  the  figure  to  an  assumed  axis 
YY',  FO  the  distance  from  center  of  gravity  to  an  axis 
XX'  at  right  angles  to  FF"  and  sum  (ax)  and  sum  (ay) 
the  sum  of  the  products  of  each  portion  of  the  figure  mul- 
tiplied by  the  distance  of  its  center  of  gravity  from  the 
F  and  X  axes  respectively. 

The  student  should  test  this  general  method  by  com- 
puting the  center  of  gravity  of  various  thin  laminas  cut 
from  cardboard  and  then  comparing  the  computed  result 

with  the  experi- 
mental result  found, 
as  in  the  succeed- 
ing article. 

An      interesting 
case     is    furnished 
by  bodies  bounded 
~\  by     curved     lines, 

J)  as    Fig.    54.     The 

— X'  area  °f  the  figure 
or  its  parts  cannot 
be  computed  readi- 
ly in  this  case.  A  result  sufficiently  good  for  practical 
use  may,  however,  be  obtained  by  dividing  the  figure 
into  narrow  elements  of  equal  width  by  lines  drawn 
parallel  to  the  FF'  axis.  Then  if  the  elements  are  suffi- 


FIG.  54 


58  ELEMENTARY  PRACTICAL  MECHANICS 

ciently  narrow,  the  length  of  the  central  line  of  each,  as 
01363  in  figure,  will  be  very  nearly  the  average  length  of  the 
element,  and  this  times  the  width  will  give  the  area.  Then, 
if  d  be  the  width  of  each  element, 


Sum 

=sum  of  areas  of  all  the  elements 
=  area  of  whole  figure. 

If  xi,  x2,  etc.,  be  distances  from   YY'  axis  to  central 
line  of  each  element, 

Sum  (dXaibiXxi+dXa2b2Xx2  +  .  .  .  )=sum  of  moments 
of  all  the  elements  about  the  YY'  axis. 

Then,  by  the  general  equation  above, 

Area  whole  figure  X  XQ  =  sum  of  moments  of  all  elements, 
or, 


_ 


Sum  (dXaibi+dXa2b2+  .  .  .  ) 


By  dividing  the  figure  into  elements  parallel  to  the  XX' 
axis,  the  value  of  YQ  may  be  found  in  a  similar  manner. 

38.  Experimental  Method  of  Finding  Center  of  Gravity. 
—  It  was  shown  in  Art.  27  that  the  center  of  gravity  of 
a  body,  suspended  from  a  single  point  in  such  a  way  that 
it  can  turn  freely  about  the  support,  must  always  be  at 
some  point  in  the  vertical  line  under  the  support.  By 
suspending  a  body  in  turn  at  two  points,  two  verticals 
containing  the  center  of  gravity  may  be  determined,  their 
intersection  therefore  gives  the  center  of  gravity  of  the 
body. 


ELEMENTARY  PRACTICAL  MECHANICS  59 


PROBLEMS 

1.  Find  force  X  for  the  system  shown  in  Fig.  55. 

2.  A  bar  AB,   14  ft.  long,  is 
pivoted  at  B.     A  weight  of   18 
Ibs.  is  hung  from  A.     Find  force 
which  must  be  applied  5  ft.  from 
A  to  produce  equilibrium. 

3.  Two   men   carry   a   weight  FIG.  55. 
of  180  Ibs.  between  them  on  a 

pole  5  ft.  long.  Where  should  the  weight  be  hung  in 
order  that  one  may  carry  three  times  as  much  of  the 
weight  as  the  other? 

4.  A  rod  AB,  the  weight  of  which  may  be  neglected,  is 
supported  2J  ft.   from  end  A.     If  the  rod  is   10  ft.   long 
and  100  Ibs.  are  hung  from  B,  what  force  will  be  required 
at    A    for    equilibrium?     What    will    be    pressure    on    the 
support? 

5.  Parallel  forces  of  75  Ibs.  and  30  Ibs.  act  in  the  same 
direction,  20  ft.  apart.     Find  amount  and  point  of  applica- 
tion of  their  resultant. 

6.  A  rod  6  ft.  long,  the  weight  of  which  may  be  neg- 
lected,  rests  upon  two  supports  placed  under  the  ends. 
Where  must  a  weight  of  22  Ibs.   be  hung  in  order  that 
pressure  on  one  support  may  be  9  Ibs.? 

7.  You  have  a  rod  2  ft.  long  and  of  negligible  weight 
and  a  balance  whose  maximum  capacity  is  64  ozs.     Show 
two  ways  in  which  you  could  arrange  your  apparatus  in 
order  to  weigh  with  this  balance  a  body  weighing  14  Ibs. 

Give  distances  on  your  diagrams 

c   and  proof  of  the  correctness  of 

your  methods. 

8.  Safety  valve -lever  AC,  Fig.      / 
56,  2  ft.  long,  and  weighing  12  \j 
FIG.  56.  Ibs.,    is    pivoted    at     A.        Its 

center    of    gravity     is     10     in. 
from  A.    Weight  of  valve  F  is  8  lbs.;  diameter  of  valve  3  in. 


60  ELEMENTARY  PRACTICAL  MECHANICS 

Valve  is  pivoted  at  J3,  4  in.  from  A.  Find  pressure  per 
square  inch  required  to  open  the  valve,  when  a  weight  W  of 
150  Ibs.  is  suspended  from  end  C  of  lever. 

9.  A  straight,  uniform  lever  AB,   12  ft.  long,  balances 
about  a  point  5  ft.  from  B,  when  weights  of  9  and  13  Ibs. 

are  suspended  from  A  and 
B,  respectively.  Find 
weight  of  lever. 


10.  AB,  Fig.  57,  is  lever 
14  ft.   long,  and   weighing 
12  Ibs.      CD  is  a  lever  12 
FIG.  57.  ft.    long    and    weighing    8 

Ibs.      Assuming  weights  of 

levers  to  act  at  their  middle  points,  what  force  can  be 
exerted  at  F  (8  ft.  from  C)  by  a  force  of  100  Ibs.  applied 
at  At 

Ak- -8^-— H«-4--*K—  -5£~ 4* — •?-'-— »1B 

11.  Find  point  of  ap-      j 1 1 1 

plication  of  resultant  of 

the  forces  in  Fig.  58.         Jt  fe      *t        J*  J* 

12.  A  uniform  beam  FIG.  58 
AB,  20  ft.  long,  weigh- 
ing 600  Ibs.,  is  supported  by  props  placed  under  its  ends. 
Four  feet  from  prop  A  a  weight  of  200  Ibs.  is  suspended, 
and  6  ft.  from  B  a  weight  of  500  Ibs.     Find  pressure  on 
each  prop. 

13.  A  rod  whose  weight  is  10  Ibs.  and  length  4  ft.  is 
supported  horizontally  by  a  smooth  peg  at  one  end  and 
a  vertical  string  15  in.  from  the  other  end.     Calculate  the 
tension  in  the  string. 

14.  Find  the  center  of  gravity  of  a  board,  8  ft.  long, 
8  in.  wide  at  one  end,  4  in.  wide  at  the  other,  tapering  equally 
on  each  side. 

15.  Compute  center  of  gravity  of  an  iron  bolt  having 
following  dimensions :  Head  f"  X£"  and  &"  thick;  threaded 
part  &"  diameter  and  2"  long. 


ELEMENTARY  PRACTICAL  MECHANICS 


61 


16.  Compute  the  pressure  on  the  head  of  the  jack,  Fig. 
12,  when  150  Ibs.  are  hung  at  C,  if  the  bar  weighs  40  Ibs. 

17.  Compute  center  of  gravity  of  the  areas  shown  in  Figs. 
59-62. 


SEi 
f.\ 


3X- 

3.  60. 


FIG.  61. 


CHAPTER  VI 


CONCURRENT  FORCES  APPLIED  AT  SEPARATE  POINTS. 
ALGEBRAIC  CONDITIONS  FOR  EQUILIBRIUM 

39.  Three  Forces  not  Parallel,  Applied  at  Different  Points 
of  a  Body. — The  graphical  solution  for  three  forces  acting 
at  the  same  point  is  given  in  Chapter  IV.  Since  the  point 
of  application  of  a  force  may  be  regarded  as  at  any  point 
in  the  line  of  action  of  the  force,  the 
same  solution  is  also  applicable  to 
cases  of  bodies  acted  upon  by 
three  forces  not  parallel  and  applied 
at  different  points  on  the  body. 

Thus  in  the  illustration  shown 
in  Fig.  63,  in  which  A B  is  any 
heavy  body  (e.g.,  a  picture,  bar, 
or  other  object),  supported  by 
flexible  cords,  AC  and  BC,  which 
are  not  parallel,  three  forces  act 
upon  AB  producing  equilibrium, 
viz.:  the  weight  W  of  AB  acting  at 
Gj  its  center  of  gravity,  and  the 
pulls  applied  by  AC  and  BC. 

The  forces  m  AC  and  BC  lie  along 
the  cords  and  therefore  their  action 
lines,  if  continued,  must  intersect  at 

some  point  C  in  the  figure.     These  forces  are,  therefore,  equiv- 
alent to  a  single  force  (their  resultant)  acting  through  (7, 

62 


FIG.  63. 


ELEMENTARY  PRACTICAL  MECHANICS  63 

and  hence  if  the  third  force  acting  on  AB  (its  weight  W], 
is  to  produce  equilibrium  with  the  forces  in  AC  and  BC, 
it  must  also  pass  through  C  and  be  equal  and  opposite  to 
the  resultant  of  the  first  two.  Or,  in  general,  no  matter 
where  on  the  body  the  forces  are  applied, 

//  a  body  acted  upon  by  three  forces  in  one  plane  and  not 
parallel  be  in  equilibrium: 

(a)  THE  ACTION  LINES  OF  THE  FORCES  MUST  ALL  PASS 
THROUGH  A  COMMON  POINT.  And 

(6)  THE  DIAGONAL  OF  THE  PARALLELOGRAM  FORMED  ON 
ANY  TWO  FORCES  AS  SIDES  AND  DRAWN  FROM  THE  POINT 
OF  CONCURRENCE  MUST  BE  EQUAL  AND  OPPOSITE  TO  THE 
THIRD  FORCE. 

In  the  case  of  a  suspended  body,  as  Fig.  63,  therefore, 
no  matter  what  the  lengths  of  the  supporting  cords  AC 
and  BC  may  be,  the  center  of  gravity  of  the  body  must 
lie  in  a  vertical  line  through  C,  the  point  of  intei  section  of 
the  cords  continued.  To  find  the  tensions  T\  and  T%, 
in  the  cords  for  a  given  weight  of  AB,  and  given  angles, 
we  may  thus  assume  the  common  point  C,  as  the  point 
of  application  of  the  forces  of  our  force  parallelogram. 

Then  to  complete  the  construction,  draw  the  vertical 
line  CR  to  represent  weight  W,  according  to  a  convenient 
scale,  and  construct  the  parallelogram  having  CR  as  a 
diagonal  and  its  adjacent  sides  in  the  directions  CT\  and 
CT2.  The  values  of  forces  T\  and  T2  are  then  determined 
by  the  lengths  of  the  lines  forming  the  sides  of  this  paral- 
lelogram. 

If  cords  AC  and  BD  are  divergent,  point  C  will  lie  below 
the  bar.  The  construction  for  tensions  in  the  cords  is 
similar  to  Fig.  63,  however,  and  should  be  made  without 
difficulty  by  the  student. 

As  a  further  illustration,  suppose  a  ladder  or  timber  AB, 
of  known  weight,  leans  against  the  vertical  side  of  a  smooth 


64 


ELEMENTARY  PRACTICAL  MECHANICS 


FIG.  64. 


building  at  any  given  angle   0  to  the  side,  as  in  Fig.  64. 

Required  to  find  the  pressure  against  the  building  and  the 

amount  and  direction  of  the 
ground  reaction  at  A.  Neg- 
lecting the  friction  at  B  against 
the  building,  the  reaction  H  of 
the  building  will  be  horizontal 
(i.e.,  there  will  be  no  vertical 
force  (friction)  to  prevent  slid- 
ing). Three  forces  therefore 
act  upon  the  ladder,  viz.:  the 
reaction  H  of  the  house,  the 
weight  W  of  the  ladder,  and  a 
force  G  equal  to  the  ground 
reaction.  H  and  W  intersect 

at  0,  hence  ground  reaction  G  must  pass  through  this  point 

for  equilibrium  (i.e.,  to  be  equal  and  opposite  to  resultant 

of  W  and  H).     In   other  words,   the   ground   both   holds 

vertically  up  on  the  ladder   (V),  and  keeps  the  end  from 

slipping  through  friction  (Fr) . 

We  now  know  the  directions  of  OP  and  OR  of  our  force 

parallelogram,  the  direction  of  the  diagonal  OQ,  and  also  the 

amount  of  force  OP  (  —  W).     Therefore,  laying  off  OP  of  a 

length  to  represent  W,  we  may 

complete  the  parallelogram  to 

scale  and  determine  force  OR= 

reaction  of  house  and  force  OQ  = 

ground  reaction.   Gis,  of  course, 

in  the  opposite  direction  to  OQ. 
A    very    common    example 

of      equilibrium     with      three 

forces  is  furnished  by  the  bent 
lever   used   in    operating 


arm 


FIG.  65. 


switches,  semaphores,  etc.,  on  railroads.     A  pull  FI,  Fig.  65, 


ELEMENTARY  PRACTICAL  MECHANICS  65 

is  applied  to  rotate  the  lever;  this  gives  rise  to  force  F2, 
which  operates  the  signal.  At  the  same  time  a  pressure 
is  set  up  against  the  pin  at  B  about  which  the  lever  rotates. 
At  any  given  position,  the  lever  is  held  in  equilibrium 
(neglecting  friction)  by  three  forces,  FI,  F2,  and  the  reac- 
tion R  of  the  pin.  Any  one  being  known  the  others  may 
be  determined.  The  construction  of  the  force  parallelo- 
gram will  be  clear  from  the  figure. 

Fig.  66  shows  a  form  of  student  laboratory  apparatus 
designed  to  illustrate  the  principle  of  the  bent  am  lever. 
The  brass  lever  with  arms  A  and  B  is  made  large  enough 
at  portion  D  to  turn  upon  ball  bearings  resting  upon  the 


FIG.  66.— Bent  Arm  Lever. 

/ 

brass  plate  C.  D  is  weighted  until  the  lever  will  rest  hori- 
zontally upon  the  ball  bearings.  The  whole  is  mounted 
upon  a  circular  board.  A  pin  is  inserted  through  the 
lever  at  P,  and  held  firmly  in  the*  plate  to  furnish  the  axis 
about  which  the  lever  may  turn. 

A  known  force  is  applied  at  F\  by  adjusting  the  clamp  F\. 
It  is  then  required  to  find: 
(a)  The  force  at  F2. 

(6)  The  amount  and  direction  of  the  pressure  on  pin  P. 

The    clamp    F2  is    adjusted,    and   the   swinging   arm  R 

rotated  about  the  board  until  the  amount  and  direction  of 


ELEMENTARY  PRACTICAL  MECHANICS 


force  at  R  is  such  that  the  lever  remains  at  rest  when  pin  P 
is  withdrawn. 

Neglecting  friction,  which  is  here  very  small  because  of 
the  ball  bearings,  three  forces  in  a  plane  are  acting  upon  the 
lever  producing  equilibrium.  (Weight  of  lever  is  here 
borne  by  reaction  of  the  support  and  hence  this  pair  of 
balanced  forces  need  not  be  considered.)  The  student  is 
expected  to  complete  the  construction,  and 

1.  Prove  that  the  unknown  pin  reaction  always  lies  along 
a  line  from  the  point  of  concurrence  of  the  action  lines  of 
the  forces  at  F\  and  F2. 

2.  Determine  required  forces  at  F2  and  R  by  means  of 
a  force  parallelogram.     These    may    then   be  checked  by 
comparison  with  the  balance  readings. 

40.  Triangle  of  Forces. — Referring  to  Fig.  39,  it  will  be 
seen  that  the  three  forces,  a,  6,  and  c,  are  equal  in  'magni- 
tude and  parallel  in  di- 
rection to  the  three  sides 
of  the  triangle  OMN.  It 
follows,  therefore,  that  if 
three  forces  in  the  same  plane 
and  acting  at  one  point  are 
in  equilibrium  they  may  be 
represented  in  magnitude 
and  in  direction  by  the  three 
sides  of  a  triangle  taken  in 


(a) 


order.  This  principle  is 
known  as  the  triangle  of 
forces.  It  is  merely  an- 
other form  of  statement  of 
the  parallelogram  of  forces 

which  is  convenient  as  a  test  for  equilibrium. 

Example  1. — In  the  simple  truss,  Fig.  41,  for  example, 

the  three  forces  acting  are  parallel    to  the  sides  of  the 


FIG.  67. 


ELEMENTARY  PRACTICAL  MECHANICS  67 

apparatus  triangle  ABC.  The  forces  bear  the  same  rela- 
tion to  each  other  therefore  as  the  sides  of  triangle  ABC. 
Or,  W: tension  in  BC: thrust  in  BA  =  CA:CB:BA. 

Example  2. — It  is  required  to  find  the  angles  at  which 
forces  of  10,  15,  and  20  Ibs.  must  act  to  produce  equilib- 
rium. First  lay  off  a  line  AB  to  represent  a  force  of  20  Ibs. 
From  A  with  a  radius  to  represent  10  Ibs.  describe  arc 
MN,  and  from  B  with  radius  to  represent  15  Ibs.  describe 
arc  OP  to  intersect  this  at  C.  Then  A  B,  BC,  and  CA 
(Fig.  67a),  represent  the  directions  of  the  forces,  and 
Fig.  676  shows  the  angles  between  them. 

41.  Polygon  of  Forces. — An  extension  of  the  triangle  of 
forces  gives  the  principle  of  the  polygon  of  forces.  Thus, 
suppose  FI,  F2,  F3,  F4,  F5  be  five  forces  acting  at  a  point 
P,  Fig.  68.  By  the  triangle  of  forces,  forces  FI,  F5,  and 
ri  will  produce  equilib- 
rium; or  7*1  is  equal  and 
opposite  to  the  resultant 
of  FI  and  F5.  Similarly, 
T2  will  produce  equilib- 
rium  with  F2  and  r\,  and 
hence  with  FI,  F2,  and 
F5;  and,  if  F4  forms  a 
closed  triangle  with  forces  FIG.  68. 

r2  and  F3,  F4  will  pro- 
duce equilibrium  with  r2  and  F3  and  therefore  with  FI,  F2, 
F3,  and  F5,  hence  the  five  forces  will   be  in  equilibrium. 
Therefore: 

//  any  number  offerees  acting  at  a  point  and  in  the  same 
plane  are  in  equilibrium  they  may  be  represented  by  the  sides 
of  a  closed  polygon  taken  in  order. 

If  force  F4  does  not  form  a  closed  triangle  with  r2,  and 
F3  (in  other  words,  if  the  five  forces  do  not  form  a  closed 
polygon),  but  lies  in  the  position  of  M  in  the  figure,  F* 


68  ELEMENTARY  PRACTICAL  MECHANICS 

will  not  produce  equilibrium  with  r%  and  F$,  and  the  five 
forces  will  not  be  in  equilibrium.  A  sixth  force,  N,  will 
be  required  for  equilibrium. 

PROBLEMS 

(Solve  Graphically) 

1.  Will  forces  of  5,  6,  and  14  Ibs.  produce  equilibrium? 

2.  Find  the  angle   at  which  the   concurrent  forces   10, 
12,  and  7  must  act  to  produce  equilibrium. 

3.  Forces  of  8,   14,   and   10  Ibs.   act  from  a  point  and 
produce  equilibrium.     Find    the    angle    between  the  8-lb. 
force  and  the  10-lb.  force. 

4.  What  horizontal  force  at  the  axle  will  be  required  to 
draw  a  wheel  of  radius  2  ft.  and  weight  20  Ibs.  over  an 
obstacle  6  in.  high?     (When  free  from  the  ground,  three 
forces  act  on   the  wheel:    Its  weight,  the  horizontal  pull, 
and  the-reaction  of  the  obstacle.) 

5.  If  AB,  Fig.  63,  is  a  uniform  metal  bar  weighing  50  Ibs., 
and  plumb   line   hanging  from   C,   the  point   of  support, 
makes  an  angle  of  30°  with  the  cord  CA,  and  an  angle 
of  40°  with  the  cord  CB,  find  the  tensions  in  CA.and  CB. 

6.  If  in  the  bent  lever,  Fig.  65,  BC  is  12  in.  long,  and  is 
at  right  angles  to  BA,  which  is  8  in.  long,  and  if  force 
Fi=S  Ibs.  is  parallel  to  BC  and  force  F2  is  parallel  to  BA, 
find   force   F%   and   the   direction   and   magnitude   of   the 
reaction  at  pin  B. 

7.  Stick  AB,  Fig.  64,  is  20  ft.  long  and  makes  an  angle 
of  70°  with  the  ground  at  A.     Its  weight  is  100  Ibs.     Find 
reactions  at  house  and  at  ground. 

42.  The  Use  of  Simple  Trigonometric  Functions. — It  is 

frequently  necessary  in  practical  mechanics  to  determine 
the  horizontal  and  vertical  components  of  a  given  force, 
to  compute  the  moment  arm  of  a  force  with  respect  to  a 


ELEMENTARY  PRACTICAL   MECHANICS  69 

given  axis,  to  determine  various  dimensions  of  a  structure 
from  other  specified  dimensions  and  angles,  etc.  These 
quantities  are  so  readily  found  through  the  use  of  simple 
trigonometric  functions  that  although  the  student  may 
have  no  previous  knowledge  of  trigonometry,  the  use  of 
such  functions  is  here  introduced.  The  necessary  ideas 
are  so  few,  so  simple,  and  so  readily  acquired,  that  time 
is  saved  in  the  end.  For  definition  of  sine,  cosine,  and 
tangent  of  an  angle,  see  Appendix. 

The  diagonal  of  any  rectangle,  as  BC,  Fig.  69,  divides 
the  rectangle  into  two  right  triangles.     By  definition, 

side  AC 

— —  =  sm  of  angle  A  BC. 

side  BC 

Therefore,  AC  =  BC  sin  angle  ABC. 

And  BC=~. -^. 

sm  angle  ABC 

If  the  length  BC  is  known  and  also  the  angle  ABC,  the 
length  of  AC  may  be  computed  by  multiplying  the  length 
BC  by  the  value  of  the  sine  of  the  angle 
ABC,  taken  from  the    table  of   sines. 
Or,  if  AC  is  known,  BC  may  be  com- 
puted by  dividing  AC  by  the  sine   of 
angle  ABC.     Similar  equations  may  be 
written  from  the  definitions  of  the  cosine  FIG.  69. 

and  tangent  of  the  angle.  All  these 
equations  are  equally  true  for  forces  AC,  BC,  etc.,  for  if 
ABC  is  a  force  triangle  we  know  from  the  parallelogram 
law  that  the  forces  are  correctly  represented  in  amount 
by  the  lengths  of  the  sides.  Further  applications  of  the 
trigonometric  method  are  shown  in  the  following  examples 


70  ELEMENTARY  PRACTICAL  MECHANICS 

Examples. — (See  Fig.  69). 

(a)  Given  £(7  =  15  and  angle  A£(7  =  40°.     Find  length 
of  AC. 

4(7  =  5(7  sin  40° 

~  =  15X.643 
-9.64. 

(See  table  of  functions,  Appendix,  p.  316.) 

(6)  Given  BA  =  25  and  angle  ABC  =  60°.     Find  BC. 

Ircos60°> 
lh500' 

5(7  =  50. 

(c)  Given  AC  =  9  and  £A  =  13.     Find  the  angle 

AC 
tan  A£(7=— • 

.O./J- 

=  _ 
13 

By  reference  to  the  table  of  tangents  we  find  that  the 
angle  whose  tangent  is  .692  is  between  34°  and  35°. 

(d)  Referring  to  Fig.  30,  suppose  force  OR  to  be  60  Ibs. 
and  angle  a,  between  its  line  of  action  and  the  horizontal, 
to  be  36°.     It  is  required  to  find  the  horizontal  and  vertical 
components  of  OR. 

Horizontal  component  OX  =  OR  cos  36  =  60  X. 809  =  48.5  Ibs. 

Vertical  component       OY  =  XR  =  OR  sin  36  =  60  X  .588 

=  35.3  Ibs. 


ELEMENTARY  PRACTICAL  MECHANICS  71 

EXERCISES 

(Solve  by  Use  of  Trigonometry) 

1.  A  force  of  50  Ibs.  acts  in  a  direction  inclined  20°  to 
the  horizontal.     Find  values  of  its  X  and  Y  components. 

2.  A  pole  20  ft.  long  leans  against  the  vertical  side  of  a 
house.      If  the  pole   makes  an  angle  of  24°  with  side  of 
house : 

(a)  How  far  from  the  foot  of  the  pole  to  the  house? 
(6)  How  far  up  the  side   of   the   house    does  the  pole 
touch? 

3.  A  horse  pulls  with  a  force  of  700  Ibs.  on  traces  making 
an  angle  of  10°  with  the  horizontal.     What  is  the  force 
pulling  the  load  horizontally,  and  what  is  the  lifting  force 
upon  the  load? 

4.  A  ladder  leans  against  a  house  with  its  foot  8  ft.  from 
the  house.     The  angle  between  the  ladder  and  the  ground 
is  75°.     How  long  is  the  ladder? 

'5.  In  a  pillar  crane,  such  as  shown  in  Fig.  74,  jib  AB 
is  16  ft.  long.  What  is  the  height  of  the  mast  if,  when 
angle  CAB  is  40°,  tie  BC  is  horizontal?  What  is  length 
of  tie  in  this  position? 

6.  A  safe  weighing  9000  Ibs.  is  being  rolled  up  planks 
which  are  inclined  36°  with  the  horizontal.  What  is  the 
perpendicular  pressure  of  the  safe  on  the  planks?  What 
is  the  force  with  which  it  tends  to  roll  down  the  planks? 

43.  Composition  of  Forces  by  Use  of  Trigonometry. — 

Instead  of  the  graphical  determination  of  the  horizontal 
and  vertical  components  of  a  system  of  forces  by  aid  of 
squared  paper,  as  in  Art.  23,  we  may  compute  such  com- 
ponents by  aid  of  the  functions  of  the  given  angles.  Thus, 
referring  to  Fig.  32, 


72  ELEMENTARY  PRACTICAL  MECHANICS 

Horizontal  component  FI  =  32  cos  20°  =  32  X. 940  =  30.1  Ibs. 

Vertical  component       F1  =  32  sin  20    =  32  X  .342  =  10.9  Ibs., 
etc. 

These  components  may  then  be  combined  just  as  before 
for  Fig.  32.  The  table  of  functions  given  in  the  Appendix 
is  correct  to  the  nearest  unit  in  the  third  place,  therefore 
computed  components  may  be  assumed  to  be  correct  to 
three  significant  figures.  This  is  a  more  rapid,  and  also  a 
more  correct  method,  than  is  usually  possible  where  graphi- 
cal methods  are  used  for  determining  the  values  of  the 
forces. 

44.  General  Conditions  of  Equilibrium  for  Concurrent 
Forces. — In  Art.  23  it  was  shown  that,  IF  ANY  NUMBER  OF 

FORCES  ACT  AT  A  COMMON  POINT,  AND  LIE  IN  THE  SAME 
PLANE,  WE  MAY  REPLACE  ALL  THESE  FORCES  BY  TWO  FORCES 
LYING  AT  RIGHT  ANGLES  TO  ONE  ANOTHER,  WHICH  ARE 
RESPECTIVETY  EQUAL  TO  THE  SUM  OF  ALL  THE  HORIZONTAL 
COMPONENTS  OF  THE  ORIGINAL  FORCES,  AND  THE  SUM  OF 
ALL  THE  VERTICAL  COMPONENTS  OF  THE  ORIGINAL  FORCES. 

In  other  words,  we  may  replace  the  original  forces  by 

Sum  X  and  sum  Y,  where  sum  X  =  algebraic  sum  of 
all  horizontal  components,  sum  Y  —  algebraic  sum 
of  all  vertical  components. 

It  is  evident  from  Fig.  33,  that  if  sum  X  =  Q  and  sum  F  =  0, 
OR,  the  resultant  of  the  whole  system  of  forces,  must  be 
zero.  The  forces  would  then  have  no  effect  upon  the 
motion  of  the  body  upon  which  they  act,  and  the  body 
would  be  in  equilibrium.  Hence, 

A  BODY  ACTED  UPON  BY  ANY  NUMBER  OF  FORCES  ACTING 
AT  ONE  POINT  AND  IN  THE  SAME  PLANE  WILL  BE  IN  EQUI- 


ELEMENTARY  PRACTICAL  MECHANICS  73 

LIBRIUM    WHEN,    IF    THE    FORCES    BE    RESOLVED    INTO    THEIR 
X   AND    Y  COMPONENTS, 


(1) 

AND  (2)  SUM  F  =  0. 

45.  Analytical  Method  of  Solution  for  Forces  in  Simple 
Structures.  —  The  conditions  of  equilibrium  just  stated 
furnish  the  simplest  method  for  the  determination  of 
stresses  in  the  members  of  simple  jointed  structures.  Such 
simple  structures  are  made  up  of  tension  and  compression 
members,  riveted  or  pinned  together  at  the  joints.  The 
forces  lie  along  the  axis  of  the  members,  and  in  cases  where 
the  weight  of  the  parts  is  included,  a  half  weight  of  each 
member  entering  a  joint  is  to  be  added  to  the  load  at 
that  point.  As  examples  of  such  solutions  the  following 
typical  cases  are  given  in  detail: 

Example  1.  —  A  body  weighing  100  Ibs.  is  being  moved  at  a 
uniform  speed  along  a  rough  horizontal  surface  by  a  force 
of  40  Ibs.  acting  at  an  angle  of  30° 
with  the  surface.  Required  the 
force  of  friction  and  the  vertical 
pressure  on  the  table. 

Consider  the  body  as  a  "  free 
body,"  i.e.,  table  removed  and  its 
place  taken  by  R  its  vertical  re- 
action, and  Fr  the  friction.  (See 

Fig.  70.)  Resolve  force  P  into  its  horizontal  and  vertical 
components  H  and  V.  Then,  by  the  conditions  of  equili- 
brium, 

(1)  H-Fr 


(2)  JR+7-(?  =  0.     (Sum  F  =  0.) 

Substituting  the  known  value  for  G,  and  the  values  for 


74 


ELEMENTARY  PRACTICAL  MECHANICS 


H  and  V  obtained  by  resolving  the  force  P,  the  unknown 
forces  Fr  and  P  may  be  found. 

Thus,   #  =  40  cos  30°,   and   F  =  40  sin  30°,   and  we  may 
write  the  equivalent  equations: 


40  cos30°-Fr  =  0. 
r  =  34.6  Ibs.  Friction. 
R+ 40  sin  30°  -100  =  0. 


(3) 
From  which, 

(4) 
From  which, 

72  =  100-20=80  Ibs.,  vertical  reaction  of  the  table. 

Example  2. — In  the  simple  truss,  Fig.  71,  tension  T 
in  tie  CB  may  be  resolved  into  its  vertical  and  horizontal 
components  V  and  H,  where  V=TsmO  and  H  =  Tcosd. 
Then,  from  the  conditions  of  equilibrium  for  the  pin  at  B 
regarded  as  a  free  body,  we  may  write  as  our  equations: 


(1) 
(2) 


Compression  R  in  AB  —  T  cos  0  =  0. 


Load  W  being  given  and  the  angle  6,  equation  (2)  may 

be  solved  directly,  and  the 
value  for  T  then  found  sub- 
stituted in    (1)   to  find  R, 
the  compression  in  the  stick. 
j*    This  latter  is  the  force  with 
j      which    A  B    pushes    to    the 
^jB    right  against  the  pin  B}  and 
also  to  the  left  against  the 
wall.     In    the   model,    the 
tension   T  in  the  tie   may 
be  checked  from  the  read- 
ing of  the  balance  in  cord 
BC.    The  compression  R  may  be  checked  by  attaching  a 


ELEMENTARY  PRACTICAL  MECHANICS 


75 


second  balance  at  B  and  pulling  horizontally  outward  until 
A  is  just  free  from  the  wall. 

Example   3. — Where   two   of  the   members   entering   a 
point  lie  outside  the  X  or  Y  axes,  i.e.,  are  neither  horizontal 


FIG.  72. 


nor  vertical,  the  student  will  find  that  his  solution  involves 

the  use  of  simultaneous  equations.     These,  however,  have 

numerical  coefficients  and  should  introduce  no  particular 

difficulties.     Thus,    in    the    structure    shown    in    Fig.    72, 

suppose   wre   know    necessary 

angles,  dimensions,  etc.,  and 

are  given  the  load   at  B,  to 

find  the  stresses  in  CB  and 

AB,  all  joints    being    free  to 

move.     Consider  the  pin  at  B 

to  be  the  "  free  body/7     We 

may  then  construct  the  force 

diagram   shown    in    Fig.    73, 

where   T  represents  the  pull  j 

exerted  by  tie  AB  and  R  the  FIG.  73.. 

thrust  of  stick  CB. 

(If  the  student  has  any  difficulty  in  understanding  why  T 
and  R  are  given  the  directions  they  here  have,  let  him 
consider:  1st.  The  effect  upon  end  B  of  apparatus  if  AB 


tsooo* 


76  ELEMENTARY  PRACTICAL  MECHANICS 

be  cut;  2d.  the  effect  if  A  B  be  replaced  and  then  CB 
be  cut.  In  what  direction  is  each  force  acting  upon  pin  J9?) 

This  is,  therefore,  a  case  of  three  concurrent  forces, 
and  is  capable  of  graphical  solution  by  direct  application 
of  the  parallelogram  law,  as  in  preceding  problems. 

Or,  if  we  resolve  R  and  T  into  their  horizontal  and 
vertical  components  and  write  the  sum  X  and  sum  Y 
equations  in  accordance  with  our  conditions  for  equilibrium, 
we  shall  have 

(1)  #-0  =  0. 

(2)  M-P-2000  =  0. 
But  Ar  =  #cos70°  =  .342#, 

M  =  Rsm  70°  =  .9407?, 


Substituting  these  values  in  (1)  and  (2),  we  have  the 
equations: 

(3)  .342£-.766T  =  0. 

(4)  .94072  -  .643  T  -  2000  =  0. 

Solving  the  simultaneous  equations  (3)  and  (4)  we 
obtain  values, 

T  =  13701bs.,     #  =  30701bs. 

Example  4-  —  In  the  two-ton  pillar  crane,  Fig.  74,  suppose 
that  a  load  of  3000  Ibs.  is  being  lifted  at  L.  Required  to 
find  tension  in  BC  and  compression  in  BA,  when  the  angles 


ELEMENTARY  PRACTICAL  MECHANICS 


77 


are  as  indicated.     Suppose  the  weight  of  jib  AB  is  1200  Ibs., 
and  that  its  center  of  gravity  is  J  of  the  way  from  A] 


FIG.  74.— Two-ton  Hand  Pillar  Crane. 

this  gives  a  vertical  load  at  5  =  ^X1200  =  400  Ibs.,  and  at 

A  =  |of    1200=800  Ibs.      The 

force  diagram   for  the   pin   at 

5,   taken   as   a    free  body,    is 

therefore    as    in    Fig.    75,   the 

letters    in    the    force    diagram  _       __j* \ 


corresponding  with  the  mem- 
bers of  the  crane  acting  at  B. 
Friction  of  the  pulley  at  B  is 
neglected  and  tension  in  part 
BD  of  hoisting  chain  is  as- 
sumed to  be  the  same  as  ten- 
sion in  part  BL. 

From   our  force   diagram,   using  letters  BC,   BA,   etc., 
to  represent  forces  in  the  corresponding  members, 


FIG.  75. 


78 


ELEMENTARY  PRACTICAL  MECHANICS 


(1)  BC  cos  20° +  3000  cos  45°  -BA  cos  45°  =  0.  (Sum  Z  =  0.) 

(2)  BC  sin  20°  +  BA  sin  45°  -  3400  -  3000  sin  45°  =  0. 

(Sum  y  =  0.) 

Substituting  values  for  the  functions  of  the  angles  and 
transposing, 


.940£C  -  .707 B A  =  -  2120. 


(3) 
(4) 


Adding  equations  (3)  and  (4), 
1.28BC  =  3400,  from  which  tension  in  BC  =  -  r  =  2660  Ibs. 

Substituting  in  equation  (4), 

.7075  A  =  5520  -  .342  X  2660,      from 
which  compression  in  B  A  =6520  Ibs. 

Fig.  76  shows  a  model  of  a  small 
laboratory  hoisting  -  crane.  The 
angles  which  jib  A  B  and  cords  BD 
and  BC  make  with  the  wall  may  be 
adjusted  as  desired,  and  sufficient 
load  applied  at  L  to  give  nearly  a 
full  reading  of  the  balance  in  BC. 
The  necessary  angles  may  then  be 
measured,  and  the  tension  in  BC  and 
compression  in  BA  computed,  allow- 
ance being  made  for  the  weight  of 
stick  BA.  The  computed  value  of 
the  tension  in  BC  may  then  be 
checked  by  the  reading  of  the  spring 
balance  in  the  cord,  and  the  computed  compression  in 


FIG.  76. — Laboratory 
Model  Hoisting-crane. 


ELEMENTARY  PRACTICAL   MECHANICS 


79 


AB  by  attaching  a  spring  balance  at  B  and  pulling  in  the 
line  AB  until  A  is  just  free  from  the  pin  at  A. 

46.  Equilibrium   when   the   Forces   Act  at   a    Common 
Point  but   do  not  Lie    in  Same  Plane.  —  Let  F\,  F2,  etc., 
Fig.    77,   be   any   number   of   given   forces   acting   at   the 
point  0.     Through  0  draw 
three  rectangular  axes,  X, 
Y,  and  Z.      It   is  evident 
that  we  may  resolve  force 
FI  into  its  separate  effects 
in   the  directions  of   these 
axes,    as   OXi,    OYi,    and 
OZ±.     In  the  same  way  F% 
may    be   replaced    by    its 
components  OX2,  OF2?  and 
OZ2,  and  any  other  forces 

FIG.  77. 


Y 


acting    at    0   may    be  re- 
placed  by  their  rectangular 

components.  We  now  have  a  new  system  of  forces  equiva- 
lent to  the  original  forces  in  which  all  of  the  forces  act  along 
the  axes  X,  Y,  and  Z.  The  resultant  of  all  the  forces 
along  the  X  axis  is  evidently  the  algebraic  sum  of  all  the 
X  components,  or  sum  X.  Similarly,  the  resultant  of 
all  the  Y  components  is  sum  Y,  and  the  resultant  of  all 
the  Z  components,  sum  Z.  Sum  X,  sum  Y,  and  sum 
Z  are  therefore  equivalent  to  the  original  forces  F\,  F2, 
etc. 

In  order  that  forces  FI,  F2,  etc.,  may  produce  equilibrium 
their  combined  effect  (i.e.,  their  resultant),  must  be  zero. 
This  is  possible  only  when  sum  X}  sum  Y;  and  sum  Z  each 
are  equal  to  zero.  Therefore: 

In  order  that  any  number  of  forces  acting  at  a  point,  but 
not  lying  in  one  plane,  may  produce  equilibrium,  it  is  neces- 
sary that: 


SO  ELEMENTARY  PRACTICAL  MECHANICS 

1.  SumX  =  0. 

2.  Sum  Y  =  0. 

3.  Sum  Z  =  0. 

PROBLEMS 

(Solve  by  application  of  the  conditions   sum  X  =  0,  sum  F  =  0, 
for  equilibrium.) 

1.  A  force  of  120  Ibs.  acting  upward  at  an  angle  of  20° 
to  the  horizontal  is  just  able  to   keep  a  body  weighing 
500  Ibs.  moving  at  uniform  speed    along  a  level    surface. 
Find  the  force  of  friction  and  the  pressure  of  the  body 
on  the  plane. 

2.  A  pull  F  of  20  Ibs.  at  an  angle  15°  to  the  plane  AB 
is  just  sufficient  to  hold  the  roller  in  an  apparatus  similar 
to  Fig.  42.     Plane  AB  is  inclined  at  an  angle  of  50°.     Com- 
pute the  weight  of  roller  and  the  pressure  perpendicular 
to  the  plane. 

3.  A   body   weighing   500  Ibs.   suspended   from   a   rope, 
is  pulled  aside  until  the  suspension  rope  is   10°  out   of 
vertical.     What  will  now  be  the  tension  in  the  supporting 
rope?      What  horizontal  pull  is  required  to  hold  the  body 
at  this  angle? 

4.  What  will  be  the  forces  in  AB  and  BC,  Fig.  72,  if, 
in  addition  to  the   2000  Ibs.   load,   there  is   a  horizontal 
pull  to  the  right  at  B  of  750  Ibs.? 

5.  Compute  the  tension  in  BC  and  compression  in  BA 
of  pillar  crane,  Fig.  74,  when  load  L  is  1000  Ibs. 

6.  Find  the  forces  in  the  members  of  the  hoisting-crane 
model,  Fig.  76,  when  L  =  30  Ibs.,  angle  BAC  =  4:Q°,  angle 
C£L  =  120°,  and  angle  CBD  =  50°. 

7.  Compute  the  forces  in  the  members  of  the  truss,  Fig.  2, 
for  a  load  of  640  Ibs.,  if  the  members  AB  and  BC  are  in- 
clined 30°  with  the  horizontal  and  if  each  weighs   10  Ibs. 
What   will  be   vertical   and   horizontal   reactions    at    foot 
of  each  stick  under  these  conditions? 


ELEMENTARY  PRACTICAL  MECHANICS 


81 


47.  Non-concurrent  Forces. — Forces  that  act  along  lines 
which  are  not  parallel  and  which  do  not  all  pass  through  a 
common  point  are  called  non-concurrent  forces. 

48.  Effects    of    Non-concurrent    Forces. — A    system    of 
oblique  forces   acting  in  lines  which   all  pass  through   a 
common  point  can   be  replaced  by  a  single  resultant,  and 
will  necessarily  produce  equilibrium  when  this  resultant  is 
zero.      The    two    conditions,   sum  Z  =  0    and    sum   F  =  0, 
where    sum  X    and    sum  Y  represent  the    algebraic  sum 


FIG.  78. 

of  all  the  horizontal  and  all  the  vertical  components  re- 
spectively, are  therefore  sufficient  for  equilibrium. 

A  system  of  non -concurrent  forces  may  produce  both  dis- 
placement and  rotation.  Such  a  system,  therefore,  often 
cannot  be  replaced  by  a  single  force.  A  general  statement 
of  conditions,  which  may  be  applied  to  all  systems  of  non- 
concurrent  forces,  must  provide  that  both  any  resultant  force 
which  may  produce  displacement  and  any  resultant  couple 
which  may  produce  rotation  shall  be  zero. 

49.  Conditions  of  Equilibrium  for  Non-concurrent  Forces 
in  One  Plane. — It  is  evident  that  we  may  resolve  the 


82 


ELEMENTARY  PRACTICAL  MECHANICS 


forces  a,  b,  c,  and  d,  Fig.  78,  into  horizontal  and  vertical 
components,  and  that  if  Hi+H2  —  H3  —  H4:  =  Q,  there  will 
be  no  tendency  for  displacement  in  a  horizontal  direction; 
if  Vi  —  V^  —  V%  +  V^  =  0,  there  will  be  no  tendency  for  dis- 
placement in  a  vertical  direction. 

We  may  also  determine  the  perpendicular  distance  from 
each  force  to  any  point  in  the  plane  of  the  forces,  as  P,  and 
compute  the  moments  of  the  forces  with  respect  to  this 
point.  Evidently  if  the  clockwise  moments  equal  the 
counter-clockwise,  or,  in  other  words,  if 


there  will  be  no  tendency  for  rotation  about  P.  And 
since  there  is  no  displacement  and  no  rotation,  the  forces 
are  in  equilibrium,  hence  we  may  state  our  conditions  for 
equilibrium  as  follows: 

A  body  acted  upon  by  a  system  of  non-  concurrent  forces 
in  one  plane  will  be  in  equilibrium  when 


1.  SumX  =  0; 

2.  Sum  Y  =  0; 

3.  Sum  of  all  moments=0. 

Fig.  79  shows  a  con- 
venient form  of  laboratory 
apparatus  for  testing  the 
conditions  of  equilibrium 
for  non-concurrent  forces. 
After  taking  the  necessary 
FIG.  79.— Laboratory  Apparatus  for  baiance  readings  dis- 

mi         t    n  i'i'  f    -r~\          'ft       •  tJC\il<A>ll\J*u  1  CCl/Lllll.i;.k5.  VAlO™1 

lest  oi  Conditions  01  Ji<quilibrium. 

tances,    etc.,    and   noting 

the  application  of  equations  (1),  (2),  and  (3),  to  the  appa- 
ratus as  here  arranged,  balances  Xi,  X%,  and  Y  should  be 
moved  so  that  the  forces  do  not  lie  in  the  X  and  Y  axes 


ELEMENTARY  PRACTICAL  MECHANICS 


83 


and  the  exercises  repeated.    The  equation  of  moments  should 

also  be  written  for  several  different  points  along  the  bar 

assumed  as  the  axis  of  rota- 

tion, and  the  fact  noted  that 

equation    (3)  is   true  for  any 

point    in    the   plane    of    the 

forces. 

50.  Solutions  of  Typical 
Cases  —  Forces  Non-concur- 
rent. —  Example  1.  —  AB,  Fig. 
80,  is  a  ladder  leaning  against 
a  smooth  wall  making  an  angle 
of  37°  with  the  wall.  The  lad- 
der is  24  ft.  long  and  its  weight 
100  Ibs.,  acts  at'D,  10  ft.  from 
B,  and  a  load  L  =  300  Ibs.  acts 
at  C,  18  ft.  from  B.  Find 

vertical  and  horizontal  reactions  at  B  and  the  horizontal 
reaction  of  the  wall  at  A.  (Since  the  wall  is  smooth 
there  can  be  no  vertical  reaction  at  A.) 


FIG.  80. 


Equations:      (1) 

(2)         Vi-W-L  =  0. 
Moments  about  B, 

(3)  100  X  10  sin  37°  +  300  X  18  sin  37°  -H2X  24  cos  37°  =  0; 
602  +  3250-  19.2#2  =  0; 
#2  =  201  Ibs. 

Therefore:  HI  =  friction,  etc.,  to  keep  ladder  irom  slipping 
=  201  Ibs.; 

7i=  vertical  component  of  reaction  of  ground 
-400  Ibs. 


84  ELEMENTARY  PRACTICAL  MECHANICS 

The  total  ground  reaction  at  B  is,  therefore, 

V(201)2+(400)2  =  4471bs.  in  the  direction  G,  making 

an  angle  with  the  ground  whose  tangent  is  -  —  =2.00. 

^U  1 

The    angle    is    therefore    approximately    63°.     (Note    that 

the  action  and  reaction  at  B  is  riot  along  ladder  AB.     Why?) 

Example  2.  —  In  the  wall  crane,  Fig.  81,  suppose  AB  =  6  ft., 

AC  =  3  ft.,   arid   load   L   is   applied   2   ft.    from   B.     Then 

_  /> 

CJ5  =  V36  +  9-6.7  ft.  approx.     Therefore  cos#  =  —  ,  sin  0 

3 

*=  —  ,  and  d  =  6  sin  6. 

The  bar  AB  is  held   in  equilibrium  by  the  tension  in 
BCj  the  reactions  V  and  H  at  the  wall  and  the  load  L. 

Therefore 

(1)  H-CBcosO  =  Q', 

(2)  V  +  CB  sintf-L  =  0; 
(3) 


If  load  L  is  known, 
CB  may  be  determined 
from  equation  (3).  By 
substituting  the  value  for 
CB  in  equations  (1)  and 
(2),  the  values  of  the 
FIG.  81.—  Wall  or  Post  Crane.  hinge  reactions  at  the 

wall  may  be  found. 

61.  Summary  of  Conditions  for  Equilibrium.  —  The  con- 
ditions for  equilibrium  under  the  action  of  the  various 
combinations  of  forces  in  one  plane  may  be  briefly  sum- 
marized as  follows: 


ELEMENTARY  PRACTICAL  MECHANICS  85 

I.  When  two  forces  are  acting:    Forces  must  be  equal  and 

opposite. 

II.  When  three  forces  are  acting:    Forces  must  be  concur- 
rent or  parallel. 

1.  If  concurrent: 

(a)  Graphical. — The  diagonal,  drawn  from  the 
point  of  concurrence,  of  the  force  parallel- 
ogram having  any  two  forces  as  adjacent 
sides,  must  be  equal  and  opposite  to  the 
third  force. 

'(&)  Algebraic. — 1st.  Algebraic  sum  of  X  com- 
ponents of  all  forces  =  0.  2d.  Algebraic 
sum  of  Y  components  of  all  forces  =  0. 

2.  If  parallel: 

1st.    Algebraic  sum  of  forces  =  0. 
2d.  Algebraic   sum  of  moments   of  all  forces 
about  any   point  in   the  plane   of  the 
forces  =  0. 

III.  When  more  than  three  forces  are  acting:   Forces  may 
be  concurrent,  parallel,  or  non-concurrent. 

1.  If  concurrent: 

(a)  Graphical. — Forces  must  be  capable  of  being 
represented  to  scale  and  in  direction  by 
the  sides  of  a  closed  polygon  taken  in 
order. 

(6)  Algebraic. — As  for  three  forces.  [See  II,  1 
(6),  above.] 

2.  If  parallel: 

Conditions  same  as  with  three  forces. 

3.  If  non-concurrent: 

1st.  Algebraic  sum  X  components  of  all  forces  =  0. 

2d.    Algebraic  sum  Y  components  of  all  forces  =  0. 

3d.    Algebraic  sum  moments  of  all  forces  about 

any  point  in  the  plane  of  the  forces^=0.-. 


86  ELEMENTARY  PRACTICAL  MECHANICS 

In  general,  for  any  combination,  equilibrium  requires 
that  there  shall  be  no  unbalanced  force  to  produce  change 
in  translation,  and  no  unbalanced  moment  to  produce 
change  in  rotation.  The  conditions  stated  under  III,  3, 
must,  therefore,  be  true  in  all  cases.  Concurrent  forces 
are,  from  this  point  of  view,  a  special  case  in  which,  as  the 
forces  can  have  but  a  single  resultant,  if  any,  the  1st  and  2d 
conditions  for  non-concurrent  forces  are  sufficient,  as  the 
3d  reduces  to  the  form  0  =  0.  Likewise  parallel  forces  are 
a  special  case  in  which  summation  along  one  axis  only  is 
possible. 

The  conditions  here  outlined  (provided  we  bear  in  mind 
that  we  may  regard  a  whole  structure  or  any  selected  part 
of  a  structure  as  the  body  in  equilibrium  under  the  action 
of  the  external  forces  applied  to  it,  and  that  the  equation  of 
moments  may  be  written  for  any  axis  through  the  plane 
of  the  forces)  furnish  all  the  equations  needed  for  the  com- 
putation of  the  stresses  in  any  simple  structure  made  up 
of  separate  members  pinned  together  at  the  points  or 
with  riveted  joints  of  small  area  which  may  be  assumed 
to  be  practically  equivalent  to  pin-connections. 


PROBLEMS 

1.  AC,  Fig  82,  is  a  uniform  steel  beam,  12  ft.  long  weigh- 
ing 20  Ibs. 

(a)  Take  moments  about  A  and  thus  find  F2. 

(b)  Take  moments  about  C  and  thus  find  FI. 

(c)  Take  moments  about  E  and  thus  find  F3. 

(d)  Resolve  F2  [found  in  (a)]  into  vertical  and  hori- 

zontal components  and  see  if  the  three  equa- 
tions for  non-concurrent  forces  are  satisfied  by 
the  numerical  values  now  found. 

2.  A  door,  Fig.  83,  is  hung  on  hinges  at  B  and  C  6  ft. 
apart;    the  door  weighs  80  Ibs.,  and  its  center  of  gravity 


ELEMENTARY  PRACTICAL  MECHANICS 


87 


A  is  1.5  ft.  out  from  the  line  passing  through  the  hinges. 
Find  vertical  and  horizontal  components  at  each  hinge. 


C     H2 


FIG.  82. 


w 
FIG.  83. 


3.  If  friction  of  ladder  against  the  building  in  Fig.  80 
is  50  Ibs.,  find  horizontal  reaction  at  A,  and  amount  and 
direction  of  ground  reaction  at 

B. 

4.  Find    tension    in   BC    and 
hinge  reactions  for  the  wall  crane, 
Fig.  81,  when  a  load  of  400  Ibs.  is 
applied  successively  1  ft.,  2  ft., 
and  4  ft.  from  A.     [See  example.] 

5.  Find  tension  in  BD  and  pin 

reactions  at  A  for  the  pole  and  pIG  g^ 

tie  shown  in  the  diagram,  Fig.  84, 

when  load  L  =  350  Ibs.,  AB  =  5  ft.,  £(7  =  1  ft.     Weight  of 

pole  =  80  Ibs.,  center  of  gravity  2J  ft.  from  A. 


CHAPTER  VII 
FORCES  IN  SOME  COMMON  COMMERCIAL  STRUCTURES 

52.  Cranes  and  Derricks. — Fig.  85  shows  a  model  of  a 
derrick  with  guy-ropes,  suitable  for  laboratory  tests. 
Spring  balances  are  inserted  in  the  tie  and  guy-ropes  to 
check  the  computed  tensions  in  these  members,  and  bal- 
ances are  also  arranged  to  check  the  reactions  at  the  foot 
of  the  boom.  The  parts  are  purposely  made  heavy  so  that 
the  effect  of  their  weight  on  the  distribution  of  forces  in 
the  members  may  be  studied,  and  provision  is  made  for 
changing  the  angles  of  the  apparatus,  and  the  points  of 
application  of  the  tie  and  load  as  desired. 

In  the  computation  of  stresses  load  L  and  weight  W  of 
boom,  acting  at  its  center  of  gravity  G,  are  assumed  as 
the  only  known  forces,  and  angles  and  distances  are  meas- 
ured as  required.  From  our  conditions  for  equilibrium, 
taking  boom  AC  as  the  free  body, 

(1) 

(2) 
(3) 

These  equations  give  the  tension  in  BD  and  the  reactions 
at  A.  The  computed  values  may  be  checked  by  the  balance 
readings.  Balances  V  and  H  should  be  read  when  the  foot 
of  boom  A  is  held  just  free  from  the  mast  by  a  horizontal 
pull. 


ELEMENTARY  PRACTICAL  MECHANICS 


89 


At  E  the  forces  are  not  in  one  plane.  We  may  imagine 
the  guy-ropes  EK  and  EF  replaced  by  a  single  guy  EM 
in  the  plane  of  EK  and  EF  and  also  in  the  vertical  plane 


FIG.  85.— Laboratory  Model  Derrick  and  Guys. 

With  the  axis  of  the  mast  and  the  load  L.  This  single  tie 
will  support  the  mast,  provided  the  boom  does  not  swing 
out  of  the  plane  of  the  tie  and  mast.  Therefore,  taking 


90 


ELEMENTARY  PRACTICAL  MECHANICS 


moments  for  the  whole  derrick  about  0,  when  there  is  no 
pull  at  Ay  we  have  the  equation, 

EMxd3-Wxd1-Lxd2--=0, 


FIG.  86.— 200  Pound  Portable  Crane. 

from    which    the    tension    in    the    imaginary    guy    EM 
may  be  found.     This  tension  may  then  be  resolved  into 

its   components   along    EF    and 
EK. 

The  reaction  R  at  the  foot  of 
the  mast  is  obviously  equal  to 
the  weight  of  whole  crane  +  L 
4-  vertical  component  of  EM. 
The  reaction  S  (when  there  is 
no  horizontal  pull  applied  at  A) 
may  be  found  from  the  equation 
of  moments  about  E, 


SxOE- 


-Lxd2  = 


FIG.  87.—  Two-ton  Hand 
Pillar  Crane. 

It  is  also   equal  to  the  hori- 
zontal component  of  tension.  EM*    Why? 


ELEMENTARY  PRACTICAL  MECHANICS 


91 


The  compression  in  the  various  sections  of  the  mast 
may  be  computed  from  the  vertical  components  of  the 
forces  above  the  sections 
plus  the  weights. 

Figs.  15,  81,  86,  87, 
and  88  show  various 
types  of  commercial 
cranes  and  derricks. 

53.  The  Shear  Legs 
— Fig.  89  shows  a  model 
of  a  pair  of  shears  used 
in  the  physics  labora- 
tories at  Pratt  Institute. 

The  legs  are  about  4  ft. 
long,  of  1£  in.  oak.  The 


FIG.  88. — Twenty-ton  Barge  Derrick. 


<feet! 


89. — Laboratory  Model  Shears. 


are  tapered  and  set  in 
holes  in  the  board  'MN 
which  is  screwed  fast 
to  the  table.  Several 
quick  adjustments  of  the 
"  spread  "  are  thus  pos- 
sible. The  legs  are  joined 
at  A  by  a  hinge,  the 
pin  of  which  passes 
through  a  clevis.  By 
fastening  a  balance  to 
the  clevis  and  pulling 
up,  along  the  line  of  a 
leg,  until  the  foot  just 
clears  the  support,  the 
thrust  at  foot  of  the  leg 
may  be  determined.  The 
tie  is  of  braided  wire,  and 
by  an  easy  adjustment  of 
its  length,  the  legs  can  be 


92 


ELEMENTARY  PRACTICAL  MECHANICS 


inclined  at  any  angle  to  the  vertical. 


FIG.  90.— Traveling  Shears. 

The  thrusts  in  the  legs  may  be 
up  by  a  single  member 
in  the  plane  of  A  B  and 
AW,  running  from  R  to  A 
(see  Fig.  92).  We  now 
have  a  simple  case  of  equi- 
librium produced  by  three 
forces  meeting  at  a  com- 
mon point.  The  stresses 
in  tie  AB  and  imaginary 
leg  AR,  are  now  found 
graphically  or  by  trigo- 
nometry. Using  the  same 
methods,  the  force  AR  is 
then  resolved  into  its  two 
components  along  the  legs 
AC  and  AD.  By  adding 
to  these  values  .7  Ib.  (the 
effect  of  the  remaining 


A  spring  balance  placed 
in  the  tie  registers  the 
tension  in  that  member. 
The  method  of  solu- 
tion usually  followed  is 
as  follows:  About  a 
20-lb.  load  is  hung  at 
W.  The  legs  together 
weigh  2  Ibs.,  and  as 
one-half  of  this  may  be 
considered  as  acting  at 
A  and  the  other  half 
at  the  feet,  one  pound 
is  added  to  the  weight 
W  in  all  computations, 
imagined  to  be  taken 


FIG.  91. — Dock  Shears. 


ELEMENTARY  PRACTICAL  MECHANICS 


93 


weight  of  the  leg) ,  the  thrust  at  the  foot  is  found.  These  results 

are  then  compared  with  the  test  readings  taken  as  described. 

A  copy  of  the  results  obtained  by  two  students,  follows: 

THE  SHEARS 

A  known  weight  W  was  hung  on  the  model,  and  the 
angles  between  the  different  members  measured.  The 
compression  in  the  legs  and  the  tension  in  the  tie  A B 
were  then  computed  by  the  following  method,  and  the 
results  obtained  checked  by  reading  the  balances  placed 
in  the  line  of  tie  and  legs: 


DATA 

W •   18.4  Ibs. 

Weight  of  AC 1      " 

Weight  of  AD 1      " 

Angle  WAB 67° 

Angle  WAR 25° 

Angle  CAR 31° 

Angle  DAR 31° 

Method  of  solution: 

By  Fig.  92  it  is  seen  that  forces  AD  and  AC  may  be 
replaced  by  a  single  force  AR, 
which  will  be  their  resultant. 
Using  this  resultant  together 
with  forces  A  B  and  AW,  & 
force  diagram  (Fig.  93)  of  the 
point  A  was  drawn  in  which 
AW  represented  the  known 
weight  W',  AB  the  tension  in 
tie;  AR  the  resultant  of  the 
thrusts  in  legs  A  B  and  AC. 

[One-half     the      combined 

weight  of  legs  AB   and  AC  w 

is  considered  as  acting  at  A,  FIG.  92. 

the  other  half  at  B  and  C.] 


94 


ELEMENTARY  PRACTICAL  MECHANICS 


*(a)  To  shelve  for  forces  AB  and  AR: 

General  law  of  equilibrium  of  forces  in  one  plane 
acting  at  a  point: 


Sum  X. 

(1) 

Sum  Y. 
(2) 


Sum  X  =  0. 
Sum  F  =  0. 

AR  sin  25°  -AB  sin  67°  =  0. 


AR  cos  25°  -  AB  cos  67°  - 19.4  =  0. 
Solving  (1)  and  (2), 

Force  AR  =  26.7  Ibs., 
Force  AB  =  12.3  Ibs. 


W 

18.4 -f  1.00  * 


FIG.  93. 


FIG.  94. 


The  balance  placed  in  tie  AB  read  12.2  Ibs. 

(6)  The  force  AR  was  then  graphically  resolved  into  its 
two  components  along  the  legs  AC  and  AD  by  means  of 
the  parallelogram  law  (Fig.  94). 


ELEMENTARY  PRACTICAL  MECHANICS 


95 


Compression  in  leg  A  (7  =  15.5  Ibs. 
Compression  in  leg  AD  =  15.5  Ibs. 

The  thrust  at  the  feet  equals  15.5  +  .7  Ibs.  =  16.2  Ibs. 

The  check  reading  of  a  balance  when  a  pull  was  exerted 
at  A  to  free  AD  at  D  was  16.3  Ibs.  The  computed  results 
vary  about  1  per  cent  from  experimental  values. 


SUMMARY  OF  RESULTS 


Member. 

Tie  AB  . 
Leg  AC  . 
Leg  AD . 


Computed 
Values. 

12. 3  Ibs. 
16. 2  Ibs. 
16. 2  Ibs. 


Check 
Readings. 

12. 2  Ibs. 

16. 3  Ibs. 
16. 3  Ibs. 


Percentage  of 
Variations. 

1% 
Wo 


Figs.  90  and  91  show  commercial  shear  legs,  the  stresses 
in  the  members  of  which  for  a  given  position  and  load 
may  be  found  in  the  same  manner. 

54.  Stone  Tongs,  etc. — The  tension  in  the  chains  AB 
and  AD  of  a  pair  of  common  stone  tongs,  Fig.  95,  for  a  given 


FIG.  95.— Stone  Tongs, 

pull  Pj  and  a  given  size  and  spread  of  tongs,  may  be  ob- 
tained either  graphically  or  algebraically,  as  ring  A  is  evi- 


96 


ELEMENTARY  PRACTICAL  MECHANICS 


60 


40 


20 


dently  held  in  equilibrium    by  three    concurrent    forces. 

The  force  R  with  which  they  grip  the  block  which  they 

support    may    then    be    computed   from    the   equation    of 

moments  about  the  pin  0. 

55.  Bridge    and   Roof   Trusses.— Figs.    96,    97,    and   98 

show  typical  forms  of  simple 
trusses  as  used  in  bridge  con- 
struction, etc.  The  stresses  in 
the  members  of  trusses  of  this 
sort  may  be  found  by  taking 
,  x  each  joint  in  turn  as  a  free  body 

»  60\A  J 

and  applying  to  each  the  equa- 
tions  for  equilibrium  for  a  system 
of  concurrent  forces.  Thus,  sup- 

FIG.  96.-The  Pratt  Truss.      P0se    the   Pratt    trUSS>    FiS'    96> 

bears  loads  which  are  expressed 

in  tons  by  the  figures  placed  near  the  arrows.  Required  to 
find  the  stress  in  each  member  of  the  truss  and  to  deter- 
mine in  each  case  whether  the  member  is  under  tension 
or  compression. 


50 


40 


E*     D 


I  J  K  L  M 

FIG.  97.— The  Warren  Truss. 


H       I       J       K       L 

FIG.  98.— The  Howe  Truss. 


(a)  Pier  Reactions. — The  sum  of  the  loads  is  230  tons, 
and  since  the  truss  is  symmetrically  loaded,  the  reactions 
R  and  Rf  at  the  piers  must  each  be  115  tons. 

(b)  Forces  at  Point  A. — At  A  three  forces  act,  the  upward 
reaction  R,  now  known  to  be  115  tons,  a  force  from  mem- 
ber B  A,  which  from  inspection  we  should  expect  to  be  a 
thrust  toward  A,  and  force  from  member  AH  which  we 
will  assume  to  be  under  tension,  and  therefore  pulling  A 


ELEMENTARY  PRACTICAL  MECHANICS  97 

toward  H.  Considering  the  joint  at  A  as  a  free  body, 
and  lettering  the  forces  in  the  same  way  as  the  members, 
we  have  the  force  diagram  shown  in  Fig.  99.  Therefore 
the  sum  X  and  sum  Y  equations  are: 


BA 

115-BAsin60°  =  0. 
Solving,  B  A  =  132.8  tons 

and  AH  =  66.4  tons. 

These  equations  check  in  every  way  and  we  may  note 
that  our  first  assumptions  were  correct,  and  that  member 
AB  is  pushing  against  A  (and  therefore  also  against  B)y 
and  hence  is  under  compression;  member  AH  is  pulling 
on  A  (also  on  H)  and  is  therefore  under  tension. 

(c)  Forces  at  Point  H.  —  Next   choose  point  H  and  draw 
its  force  diagram  (Fig.  100).     Remember  if  a  member  of 
j  any  structure  is  under  tension,  it  must 

pull  equally  and  in  opposite    directions 
upon   the   two   joints   of   the   structure 
A___  between  which  it  is  attached,  or  if  under 
^     compression,    must  push  equally   upon 
the  two  joints.     Every  time  we  deter- 
mine the  stress  in  a  member  from  a 
1  consideration  of  the  conditions  at  one 

end,  we  also  determine  the  action  which 
it  exerts  upon  the  joint  at  the  other  end.  Inspection  shows 
that  for  equilibrium  HC  must  be  equal  and  opposite  to  the 
known  tension  HA  =  66  .4  tons,  and  also  the  member  HB 
must  be  pulling  up  from  H  with  a  force  equal  to  the 
40-ton  load  at  H. 

.'.     Cff  =  66.4  tons  (tension), 

40     tons  (tension). 


98          ELEMENTARY  PRACTICAL  MECHANICS 

(d)  Forces  at  Point  B. — Next  take  point  B.  Its  force 
diagram  may  be  drawn  as  in  Fig.  101.  It  is  known  from 
the  previous  solutions  that  the  member  A  B  pushes  up 

and  to   the  left   with   a   force 

m8  *  ^  (found  at  point  A)  of  132.8  tons. 

There  is  the  downward  act- 
ing load  of  20  tons  and  also  the 
tension  of  member  BH  =  4Q  tons 
(see  point  H).  There  are  two 
other  forces  acting  at  B  whose 
amounts  are  as  yet  unknown; 
one  from  member  CB,  along  the 
line  we  will  call  in  force  dia- 
gram CC',  and  the  other  from 

member  FB  along  the  line  F'F.  We  will  call  these  forces 
BC  and  BF,  but  assume  that  as  yet  we  do  not  know  which 
way  they  act;  i.e.,  whether  members  CB  and  FB  are 
compression  or  tension  members.  Assume  force  BC  to  be 
upward  and  to  the  right  (i.e.,  compression),  force  BF  to  be 
to  the  right  (compression),  and  write  the  equations  for 
point  B. 

- 132.8  cos  60 + BC  cos  60 + BF = 0,         (Sum  X = 0.) 
and 

-  40  -  20  + 132.8  sin  60 +BC  sin  60=0.     (Sum  7=0.) 

Solving  the  Sum  Y  equation  we  get  £=—  63.5  tons. 
The  negative  sign  shows  that  the  force  along  BC  must  be 
such  as  to  give  a  negative  value  for  its  Y  component,  i.e., 
must  be  in  direction  BC']  or,  in  other  words,  BC  must 
be  a  tension  member  instead  of  compression  member  as 
assumed. 

Correcting  the  sign  for  BC  (i.e.,  the  direction)  in  our 
Sum  X  equation,  and  solving  for  FB}  we  obtain 

+98.2  tons. 


ELEMENTARY  PRACTICAL  MECHANICS  99 

Here  the  positive  sign  shows  that  the  member  FB  acts 
to  the  right  at  B,  as  we  assumed  in  writing  the  equations, 
and  BF  s  therefore  a  compression  member. 

(e)  Force  at  Point  F. — Next  take  point  F.  From 
symmetry  we  know  EF  =  BF  =  98.2  tons  compression. 
The  only  unknown  is  member  CF  which  inspection  shows 
must  be  in  compression  and  equal  to  60  tons  to  balance 


FIG.  102. — Through  Span  over  the  Wabash-Pittsburg 
Terminal  Railway. 

the  load  at  F.     Since  the  other  half  of  the  truss  is  loaded 
in  the  same  way,  the  forces  in  all  the  members  are  now 
known,  but  as  a  check  we  may  figure  out  point  C. 
The  vertical  forces  are  given  by  the  equation: 

-5Q-FC  +  BC  cos  3Q  +  EC  cos  30  =  0. 
Or,       -50-60  +  2X63.5X.866=0,  because  EC  =  EC. 

This  reduces  to  -50-60  +  110  =  0,  thus  checking   all  the 
previous  calculations. 

Figs.  102  and  103  show  bridge  trusses  in  actual  use,  the 
stresses  for  which  under  given  conditions  of  load  may  be 
determined  in  the  same  manner. 


100 


ELEMENTARY  PRACTICAL  MECHANICS 


Such   structures   are   subject   to   both   a   "dead"   load, 
from  weight  of  parts,  ties,  track,  etc.,  and  to  a  varying 


FIG.  103. — Baltimore  and  Ohio  Railroad  Bridge  crossing  South 
Branch  of  Potomac  River  in  West  Virginia. 


„       „ Tic  8\  9"  lc"c.toC. 

Guard  Rail  C  x  8-^"~njL          i  ^n ru         /    Tr 


lts  through 
CoLCap 


-si 

]i 

Total  width  of  Street  80 'o*  - 


ivets  * 


nttV 


Castings 

Brie 

Bric 


k    /-foSciw 
k    ',  >      B  lt  - 

ry^j-^    /  \         - 


Live  Load 

-  Engine-- 


Concrete, 


mtet* 


r— 


FIG.  104. — Elevated  Railway,  Brooklyn,  N,  Y. 


or  "  live  "  load,   coming  from  the  weight  of  the  engine, 
cars,  loaded  wagons,  etc.,  hauled  over  them.     These  loads 


ELEMENTARY  PRACTICAL   MECHANICS 


101 


may  be  applied  at  the  joints  by  the  method  of  construc- 
tion (e.g.,  Fig.  104,  which  shows  the  manner  in  which  the 
tracks  of  the  Kings  County  Elevated  Railway  of  Brooklyn 
are  supported) ;  if  not  the  distributed  load  should  be  con- 
centrated at  the  joints  for  the  purposes  of  computation, 
just  as  in  previous  instances  we  have  allowed  for  the  weight 
of  a  part  by  assuming  a  half-weight  as  acting  at  each  end. 
Figs.  105,  106,  and  107  illustrate  typical  forms  of  roof 
trusses  which  are  also  capable  of  solution  by  the  "  point- 
by-point  "  method.  In  every 
instance  the  reactions  of  the 
end  supports  should  be  deter- 
mined first,  and  the  joint  at 
a  support  taken  as  the  first 
point.  We  may  then  proceed  .FlG  105 

from    joint    to    joint    in   any 

order  convenient  so  long  as  there  are  not  more  than  two 
unknown  forces  at  the  joint   considered.     As  we  have  only 


FIG.  106.—  Boiler  House  Roof  Truss. 


the    two    equations,    sum    X    and    sum    F  =  0,   only  two 
unknowns  are  determinate.     In  trusses  where  these  con- 


102 


ELEMENTARY  PRACTICAL  MECHANICS 


FIG.  107. 


ditions  are  not  met,  moments  must  be  taken  about  selected 
points  and  the  unknown  forces  reduced  to  two  in  number. 
The  reactions  of  the  supports  may  be  determined  from 
inspection  in  the  case  of  symmetrical  trusses  which  are 
symmetrically  loaded.  In  other  instances,  these  reactions 
must  be  computed  by  taking  moments  about  the  supports, 
as  for  parallel  forces. 

The  "  dead  "  loads  for  roof  trusses  consist  of  the  weight 
of  the  truss  and  the  weight   of  the  section  of  the  roof 

supported.  The  "  live " 
loads  are  the  wind  and 
snow  loads.  These  may 
be  applied  to  the  joints  by 
an  auxiliary  piece  in  the 
construction  of  the  .  truss 
or  in  any  case  may  be 
carried  to  the  joints  or 
apex  for  the  purposes  of  computation  as  follows:  Let 
shaded  area  MNOP  in  Fig.  108,  which  is  a  top  view  of  a 
section  of  the  roof,  represent  the  portion  of  the  roof  and 
snow  load  taken  by  the  truss 
A  EC.  This  area  will  extend 
half  way  to  the  trusses  on  either 

M 

side,  and  the  total  weight  of  roof 

and  snow  will  be  equal  to  the    A 

weight    per    square    footXarea 

MNOP.     Of  this,  the  weight  of 

the  half  area  MQRP  comes  on 

the  member  AE,  and  the  weight 

of  the  other   half   area  QNOR 

comes  on  EC.     We  may  therefore  regard  the  load  on  AE 

as  the  total  weight  W  of  area  MQRP  acting  at  the  center 

of  gravity  of  the  area,  which  is  the  middle  point  of  AE,  or 

what  is  the  same  thing,  as  acting  \  at  E  and  J  at  A.    The 


FIG.  108. 


ELEMENTARY  PRACTICAL  MECHANICS  103 

load  on  A  E  is  therefore  equivalent  to  \  W  at  E  and  £  W 
Sit  A.  In  the  same  way  load  on  EC  is  \  W  at  E  and  %  W 
at  C.  The  total  apex  load  at  E  is  therefore  %W  +  %W  =  W, 
and  the  loads  at  A  and  C  are  each  J  17.  The  student 
should  note  that  the  loads  at  A  and  C  produce  pressure 
on  the  supports  but  have  no  effect  in  producing  stress  in 
members  AE  and  EC  of  the  truss. 
66.  The  Sulkey  Derrick.— The  solution  for  the  forces  in 


FIG.  109 .—Contractor's  Sulkey          FIG.  110.— Tank  City  Water 
Derrick.  Works,  Sycamore,  111. 

the  members  of  a  "  Sulkey  "  derrick,  Fig.  109,  may  be 
made  most  readily  by  assuming  the  legs  CA  and  CB  re- 
placed by  a  single  leg  in  the  plane  of  CD  and  the  load,  as 
was  done  in  the  case  of  the  shear  legs.  The  forces  at  C 
then  constitute  a  system  of  concurrent  forces  which  may 
be  determined  in  the  usual  way.  The  compression  in  the 


104 


ELEMENTARY  PRACTICAL    MECHANICS 


single  leg  may  then  be  resolved  into  its  components  along 
the  legs  CA  and  CB. 

57.  Water  Tanks.  —  The  common  water  tank  shown  in 

Fig.  110  furnishes  another  illustration  of 
equilibrium  with  concurrent  forces  hav- 
ing components  in  three  axes  X,  Y,  and 
Z.  A  simple  solution  because  of  the 
ease  with  which  dimensions,  angles,  etc., 
may  be  obtained  is  suggested  in  Fig. 
111.  Weight  W  of  tank  and  contents 
is  supported  by  the  reactions  of  the  four 
legs.  If  continued  these  five  forces  will 
intersect  at  a  common  point  at  0.  If 
now  we  assume  forces  OA  and  OB  re- 
placed by  a  single  force  OM,  and  OC 
and  OD  by  the  single  force  OJV,  we  have  a  system  of  con- 
current forces  in  a  plane.  The  required  forces  OM  and 
ON  may  now  be  obtained  graphically  or  algebraically  in 
the  usual  way,  and  OM  and  ON  may  then  be  resolved  into 
their  components  along  OA  and  OB}  and  OC  and  OD, 
respectively.  These  are  the  compressions  in  the  support- 
ing legs. 

58.  The  Arch.  —  The  reactions  at  the  ends  and  the  com- 
pression in  the  center  of  an  arch  may  be  found  by  regarding 
one  half  of  the  arch  as  a  free  body.     The  arrangement  of 
forces  shown  in  Fig.  112  then  results,  which  is  that  of  a 
simple  system  of  non-concurrent  forces.     Therefore 


(1) 
(2) 
(3) 


-TF-TFi,  etc.,  = 


Loads  W,  Wij  etc.,  being  known  and  the  dimensions  of 


ELEMENTARY  PRACTICAL   MECHANICS 


105 


the  arch,  these  equations  may  be  readily  solved  for  7  and 
HI,  the  end  reactions,  and  H2  the  compression  at  the 
center  of  the  arch  ring. 
The  end  reactions  for  the 
other  half  are  the  same  if 
the  arch  is  loaded  uni- 
formly, if  not  similar  equa- 
tions may  also  be  written 
for  the  second  half.  Fig. 
113  shows  a  laboratory 
model  of  an  arch  in  which 
the  computed  forces  may 

be  checked  by  spring  balances.  One  end  only  of  the  arch 
is  left  free  or  "  floating  "  at  a  time,  the  other  turns  freely 
about  a  pin  through  the  horizontal  frame  which  serves  as 
a  guide  to  maintain  the  arch  in  a  vertical  position. 


FIG.  113. — Laboratory  Model  Arch. 


106  ELEMENTARY  PRACTICAL  MECHANICS 


PROBLEMS 

1.  In  the  derrick,  Fig.  15,  mast  CA  is  14J  ft.  long,  boom 
CD  20  ft.  long.     Guys  AG  and  AF  are  each  18  ft.  with 
8  ft.  spread  at  the  feet.     AE  is  4  in.  and  boom  weighs 
320  Ibs.     Find  tension  in  the  tie  and  the  guys,  and  thrust 
of  boom  at  C,  when  load  L  is  2  tons  and  boom  makes  angle 
of  45°  with  the  mast,  and  its  vertical  plane  bisects  the 
angle  GAP. 

2.  In  the  portable  crane,  Fig.  86,  center  of  gravity  of 
the  crane  and  also  of  the  base  is  directly  over  the  axle 
for  wheels  EE.     Weight  of  crane  120  Ibs.,  of  platform  90  Ibs. 
From  vertical  plane  through  axle  to  front  edge  B  of  base 
is  6  in.,  to  edge  C  is  9  in.;    to  A  is  3^  ft.,  to  front  and 
back  wheels  each  2  ft.     What  force  will  be  required  at  C 
when  200  Ibs.  are  hung  at  A  if  there  are  no  other  fasten- 
ings?    What  will  then  be  the  pressure  on  each  wheel? 


3.  In  the  pillar  crane,  Fig.  87,  AJ5  =  10  ft.,  BD  =  8±  ft., 
AC  =  7%  ft.,  CD  =  7i  ft.,  CB  =  3i  ft.,  chain  from  C  to  E  is 
vertical.     Find  stresses  in  all  members  when  load  L  is 
2  tons.     (Neglect  weight  of  parts  and  friction  and  note  that 
the  chain  runs  through  block  L  only  once.) 

4.  The  legs  of  the  shears,  Fig.  91,  are  each  80  ft.  in 
length,  with  a  spread  of  20  ft.  at  the  base,  and  a  vertical 
height  of  60  ft.     The  tie   and  hoisting  chain   both   reach 
from  the  top  of  shears  to  a  point  40  ft.  back  of  the  middle 
point  in  the  line  between  the  feet.     Find  stresses  in  the 
members  when  supporting  a  load  of  20  tons,  if  tension 
in  hoisting  chain  winding  on  the  drum  is  J  of  the  load. 

5.  Find  tension  in  chains  and  force  R  for  stone  tongs, 
Fig.  95,  when  lifting  a  block  weighing  1500  Ibs.,  if  di  =  18  in., 
d2  =  15  in.,  and  angle  BAD  =  140°. 

6.  Find   stresses   in   the   members   of  the   Howe   truss, 
Fig.  98,  when  loads  are  20  tons  at  each  joint,  at  bottom 
and  5  tons  for  each  joint  at  top.     Angle  BAL  =  6Q°   angle 


ELEMENTARY  PRACTICAL  MECHANICS  1G7 

7.  Find  stresses  in  the  members  of  truss  shown  in  Fig.  102, 
when  a  truck  weighing  10  tons  stands  on  center  of  bridge 
and  dead  loads  at  each  joint  are  %  ton.     Angles  of  truss 
as  in  Fig.  96. 

8.  In  the  roof  truss,  Fig.  105,  angles  FAB  and  ABF  are 
each  30°.     Loads  of  2  tons  at  F  and  D  and  3  tons  at  E. 
Compute  stresses  in  all  members. 

9.  Find  stresses  due  to  live  loads  in  the  members  of  the 
Warren  truss,  Fig.  97,  if  each  span  AM,  ML,  etc.,  is  8  ft., 
if  all  angles  are  60°,  and  if  live  loads  are  10  tons  at  Ky 
and  5  tons  each  at  J ,  L,  and  M. 

10.  Find  stresses  in  each  member  of  the  roof  truss  shown 
in  Fig.  107  if  loads  are  5  tons  each  at  C  and  B  and  20  tons 
at  E. 

11.  Find  forces  in  the  members  of  the  sulkey  derrick, 
Fig.  109,  due  to  load,  if  the  legs  are  each  15  ft.  long,  6  in. 
from  center  to  center  at  C,  distance  AB  =  4:  ft.,  distance  D 
to  middle  of  line  ^5  =  10  ft.,  load  L=1000  Ibs. 

12.  Find  compression  in  the  supports,  Fig.   110,  when 
tank  contains   10,000  gallons  of  water,  if  AB,  BC,  etc., 
are  each  20  ft.,  A'B',  B'C',  etc.,  each  10  ft.,  and  vertical 
height  of  legs  60  ft.     Weight  of  tank  5  tons. 

13.  Compute  horizontal  thrust  in  middle  of  arch  ring 
and  vertical  and  horizontal  reactions  at  the  piers  for  an 
arch  with  dimensions,  and  loads,  as  in  diagram,  Fig.  114. 


FIG.  114. 


CHAPTER  VIII 
MOTION 

BEFORE  beginning  this  chapter,  the  student  should  review 
the  brief  discussion  of  motion  and  velocity  given  in 
Chapter  II;  also  the  application  of  graphical  methods  and 
the  parallelogram  law  to  motions  and  velocities,  Chapters 
II  and  III.  It  should  be  understood  that  velocities,  or 
accelerations,  may  be  combined,  resolved  along  axes,  etc., 
just  as  we  have  learned  to  do  with  forces. 

59.  Speed.  Velocity. — The  rate  of  motion  of  a  body 
is  termed  its  speed. 

Thus  a  person  may  travel  with  a  speed  of  30  miles  per 
hour,  a  railroad  train  may  increase  its  speed,  etc. 

The  term  speed  expresses  only  two  ideas:  distance  passed 
over,  and  time.  It  will  be  evident,  however,  from  a  little 
consideration,  that  these  alone  do  not  furnish  sufficient 
data  for  a  full  consideration  of  a  moving  body.  Thus, 
two  trains,  one  running  on  a  straight  track,  the  other 
around  a  curve,  may  both  have  the  same  speed;  the  stresses 
to  which  the  two  are  subject,  and  the  resistance  each  offers 
to  its  motor,  will  be  very  different. 

To  account  for  these  forces,  we  must  know,  not  only  the 
speed,  but  also  the  direction  of  the  motion,  and  even  the 
rate  at  which  the  direction  is  changing,  as  well  as  the 
rate  at  which  the  speed  is  changing.  To  this  full  descrip- 
tion of  a  motion,  which  states  the  three  specifications, 
displacement,  time,  and  direction,  the  term  VELOCITY  is 
applied. 

108 


ELEMENTARY  PRACTICAL  MECHANICS  109 

This  distinction  between  the  terms  speed  and  velocity 
will  be  made  evident  by  the  following  illustration:  Sup- 
pose a  railroad  train  starts  from  a  given  station  and 
travels  for  3  hours  in  a  straight  line  at  a  speed  of  30  miles 
per  hour.  From  this  description  of  the  train's  motion,  all 
we  know  regarding  its  position  at  the  end  of  the  three  hours 
is  that  it  is  somewhere  90  miles  away  on  a  straight  line 
from  the  station.  But  suppose  the  train  moves  with  a 
velocity  of  30  miles  per  hour  due  north.  Its  position  at  the 
end  of  the  three  hours,  or  after  the  lapse  of  any  part  of  the 
time,  is  now  definitely  fixed.  A  knowledge  of  the  velocity 
of  a  body  therefore  enables  us  to  determine  its  position 
after  any  time.  Hence  velocity  may  be  defined  as  rate 
of  change  of  position. 

In  speaking  of  the  motion  of  bodies  whose  direction 
of  motion  is  definitely  recognized  as  fixed  by  the  condi- 
tions, as,  for  example,  a  falling  body,  a  part  of  a  machine, 
a  train  on  a  given  stretch  of  track,  etc.,  we  may  use  the 
terms  speed  or  velocity  without  danger  of  confusion, 
although  no  actual  specification  of  direction  is  given  for 
the  latter. 

60.  Statement    of    Velocity. — The    velocity    of    a    body 
defines  its  motion  at  any  given  instant  only.     Velocity  may 
remain   constant   in   speed   and   direction,    or  either   may 
change  from  instant  to  instant.     To  express  velocity,  we 
state    the    distance    the    body   would    move   in   a   given    (or 
recognized)   direction  in  one  unit  of  time  if  it  maintained 
unchanged  its  present  speed  and  direction;    thus:    30  miles 
(distance)   per  hour   (time),   10  ft.  per  second,  etc.     This 
statement  must  not  be  interpreted  as  meaning  that  the 
body  must  actually  move  over  the  stated  distance,  or  that 
motion  must  continue  for  the  entire  unit  of  time. 

61.  Velocity  of  Rotation.     Angular  Velocity. — The  veloc- 
ity of  a  point  on  a  rotating  body,  as,  for  example,  a  point 


110  ELEMENTARY  PRACTICAL  MECHANICS 

on  the  rim  of  a  fly-wheel,  may  be  expressed  in  two  ways: 
1st,  By  stating  the  actual  length  of  arc  swept  over  by  the 
point  in  unit  time',  or,  2d,  by  stating  the  rate  at  which  the 
point  is  changing  its  direction  from  the  axis  of  rotation. 
The  latter  expression  is  called  the  angular  velocity. 

It  is  evident  that  the  first  expression  —  i.e.,  the  actual 
linear  velocity  of  the  point  —  depends  not  only  upon  how 
fast  the  wheel  is  turning,  but  also  upon  the  distance  from 
point  to  axis.  Points  farther  out  from  the  axis  have  a 
greater  linear  velocity  than  points  situated  nearer  the  center. 
The  angular  velocity  is,  however,  the  same  for  all  points 
on  the  wheel.  Angular  velocity  may 
be  stated  therefore  for  the  rotating 
body  as  a  whole. 

Angular  velocity  may  be  expressed 
in  terms  of  degrees  turned  through 
per  unit  time,  or,  as  is  more  customary, 
in  terms  of  a  special  unit  called  a 
radian.  The  angle  d  at  the  center, 
which  intercepts  an  arc  A  B,  equal  in 
length  to  the  radius  r  of  the  circle  (Fig.  115)  is  called  a  radian. 
Since  the  circumference  of  a  circle  equals  27rr,  and  each 
radian  represents  an  arc  of  length  r,  it  is  evident  that 


there  are   —  =  2^  radians  in  a   complete   circumference; 
r 

O/^A 

and   that    one    radian  =  —=57.3   degrees  approximately. 

2?r 

The  advantage  of  this  method  of  expressing  angular  veloci- 
ties is  evident  from  the  fact  that  angular  velocity  in  radians 
Xdistance  r  from  point  to  axis  of  rotation  =  linear  velocity 
of  the  point.  Thus,  suppose  a  wheel  has  an  angular  velocity 
of  5  radians  per  second;  a  point  on  this  wheel  2  ft.  from 
the  axis  then  has  a  linear  velocity  of  5  X  2  =  10  ft.  per  second. 
In  connection  with  machinery,  velocity  of  rotation  is 


ELEMENTARY  PRACTICAL  MECHANICS  111 

commonly  expressed  in  "  revolutions  per  minute."  Revo- 
lutions per  minute  X  2?r  =  radians  per  minute. 

62.  Uniform  Motion.  Variable  Motion. — A  classification 
of  motions  of  great  importance  in  mechanics  is  that  of 
uniform  motion  and  non-uniform,  or  variable  motion.  Thus, 
if  a  runner  goes  steadily  20  ft.  every  second  and  a  pro- 
portional distance  for  any  fraction  or  multiple  of  a  second, 
his  motion  is  said  to  be  uniform;  if  a  fly-wheel  revolves 
steadily  200  times  a  minute,  its  motion  is  also  uniform. 
But  a  falling  stone,  a  train  getting  under  way  from  a 
station,  or  coming  to  a  stop,  the  piston  of  an  engine,  etc., 
are  cases  where  the  motion  is  not  uniform. 

Uniform  motion  is  motion  with  a  constant  or  unchanging 
velocity. 

Variable  motion  is  motion  where  the  velocity  is  changing 
from  instant  to  instant. 

When  the  velocity  is  variable,  it  is,  at  any  instant, 
measured  by  the  distance  through  which  the  body  would 
pass  in  one  second  if  it  were  to  continue  to  move  with  the 
velocity  it  had  at  that  particular  instant.  This  is  called  the 
instantaneous  velocity  of  the  body. 

In  uniform  motion,  whether  of  translation  or  rotation, 
the  following  relation  exists: 

DISTANCE  =  (UNIFORM)  VELOCITY  X  TIME. 
Or,  in  symbols,  S=uXt. 

PROBLEMS 

(See  problems  on  graphical  representation  of  motions  and 
velocities,  Chapter  II;  also  on  composition  and  resolution  of 
motions  and  velocities,  Chapter  III.) 

1.  Express  the  velocity  of  rotation  of  the  minute  hand 
of  a  clock  in  degrees  per  second.  In  radians  per  second. 


112  ELEMENTARY  PRACTICAL  MECHANICS 

2.  How  long  will  it  take  a  train  to  go  300  miles  with  a 
velocity  of  24  ft.  a  second? 

3.  In  what  time  will  a  body  traverse  80  kilometers  if  its 
velocity  is  to  be  200  cm.  per  second? 

4.  Which    is   the   greater   velocity,  300  revolutions   per 
minute  or  10  radians  per  second,  and  by  how  much? 

5.  A  wheel  is  rotating  with  a  uniform  speed  of  45  degrees 
per  second.     How  many  revolutions  will  it  make  in  half 
an  hour?     Express  its  velocity  in  revolutions  per  minute. 
In  radians  per  second. 

6.  Through  what  distance  will  a  point  on  the  wheel  of 
Prob.  5,  2  ft.  from  the  axis,  move  in  5  minutes? 

7.  A  wheel    is  rotating  with  a  velocity  of  2  radians  per 
second.     Through  what  distance  will  a  point  on  the  wheel 
18  inches  from  axis  move  in  1  minute? 

63.  Uniformly  Accelerated  Motion. — We  have  seen  that 
variable  motion  is  such  that  the  velocity  changes  either 
regularly  or  at  intervals.  When  the  change  of  velocity 
proceeds  uniformly,  either  increasing  or  decreasing,  the 
motion  is  called  uniformly  accelerated  motion.  This  is  the 
only  type  of  variable  motion  we  shall  need  to  consider. 

A  train  starting  from  rest  at  a  station  and  gradually 
getting  under  headway,  is  an  illustration  of  accelerated 
motion.  If  the  force  starting  the  tram  is  uniform  and 
the  resistance  to  motion  constant,  the  motion  of  the  train 
will  be  uniformly  accelerated.  The  same  train  pulling  into 
a  station  and  gradually  stopped  by  application  of  the  brakes 
is  an  illustration  of  accelerated  motion  where  the  velocity 
is  decreasing,  i.e.,  of  negatively  accelerated  or  retarded 
motion.  The  motion  of  a  body  thrown  vertically  upward 
or  falling  (freely)  from  an  elevation  under  the  influence 
of  gravity,  is  a  common  illustration  of  uniformly  accelerated 
motion.  A  shaft  starting  to  rotate  under  the  pull  of  a 
belt,  a  body  sliding  down  an  inclined  plane,  a  spinning-top 


ELEMENTARY  PRACTICAL  MECHANICS  113 

coming  to  rest,  are  all  illustrations  of  accelerated  motion; 
and  if  the  forces  causing  and  opposing  motion  are  con- 
stant throughout  the  time  considered,  they  are  cases  also 
of  uniformly  accelerated  motion. 

64.  Acceleration. — Suppose   a  body  to   start  from   rest 
and,   under  the   influence   of   a   constant   force,   to   move 
with   a  uniformly  increasing   velocity  in   a   straight  line. 
At  the  beginning  the  velocity  is  0.     Now  suppose,  in  the 
case  we  are  considering,  that  at  the  end  of  the  first  second 
the  velocity  has  become   10  ft.   a  second;    in  2  seconds, 
20  ft.  a  second;    in  3  seconds,  30  ft.  a  second,  etc.     The 
gain  in  velocity  is  10  ft.  a  second  in  each  second. 

This  gain  of  velocity  in  unit  time  is  called  the  ACCELERA- 
TION. The  body  therefore  starts  from  rest  and  moves 
with  an  acceleration  of  10  ft/sec2. 

Or,  suppose  the  body  has  a  velocity  of  30  ft/sec  at  the 
instant  we  begin  to  consider  its  motion,  and  that  one 
second  later  its  velocity  is  20  ft/sec,  two  seconds  later 
10  ft/sec,  and  three  seconds  later  0  ft/sec,  or  body  is 
at  rest;  the  loss  of  velocity,  or  retardation,  each  second 
is  here  10  ft/sec. 

It  is  customary  to  call  the  CHANGE  in  velocity  which 
occurs  in  unit  time  the  acceleration,  whether  the  change  is 
an  increase  or  a  decrease  in  velocity.  The  distinction  is 
merely  one  of  direction.  The  acceleration  is  always  in  the 
direction  of  the  force  acting  to  produce  change  of  velocity, 
but  this  force  may  be  either  in  the  direction  of  the  original 
motion  or  opposed  to  it.  In  the  first  case  the  velocity  will 
increase,  in  the  second  it  will  decrease,  i.e.,  acceleration  is 
negative  with  respect  to  the  original  velocity. 

65.  Acceleration    of   Rotating   Bodies. — Acceleration   in 
motion   of  rotation   may  be  expressed  in  the  two   ways 
already  stated  for  the  velocity.     Thus,  we  may  state  the 
change  in  Linear  velocity  per  unit  of  time,  i.e.,  the  linear 


114  ELEMENTARY  PRACTICAL  MECHANICS 

acceleration  of  a  point  at  a  known  distance  from  the  axis 
of  rotation,  or  we  may  state  the  change  in  angular  velocity 
per  unit  of  time,  i.e.,  the  angular  acceleration.  The  linear 
acceleration  of  different  points  in  a  rotating  body  whose 
velocity  is  changing  evidently  depends  upon  the  distances 
of  the  points  from  the  axis:  the  angular  acceleration  is 
the  same  for  all  points.  Angular  acceleration  may  be 
expressed  in  revolutions  per  unit  of  time  gained  or  lost 
each  unit  (e.g.,  10  revs,  per  min.  each  sec.),  or  as  radians 
per  second  gained  or  lost  each  second.*  • 


PROBLEMS 

1.  A  velocity  change  from  0  to  50  ft /sec  occurs  in  10 
seconds.     Find  the  acceleration. 

2.  If  the  acceleration  is  5  ft /sec2,  and  the  initial  veloc- 
ity is  6  ft /sec  what  will  be  the  velocity  at  the  end  of 
8  seconds? 

3.  In  what  time  will  the  velocity  of  a  body  falling  from 
rest  become  100  ft/sec  if  the  acceleration   of    gravity  is 
32  ft/sec2? 

4.  If  at  1  o'clock  the  velocity  of  a  train  is  15  miles  per 
hour  and  at  2  minutes  past  1  is  20  miles  per  hour,  what 
is  the  acceleration  in  miles  per  hour  per  minute? 

5.  What  is  the   acceleration   in   Prob.   4,   expressed  in 
miles  per  minute  per  minute? 

6.  A  certain  body  moves  with  an  acceleration  of  80  ft /sec2. 
What  will  be  the  change  in  velocity  in  half  a  minute? 

7.  Express  the  acceleration  in  Prob.  6  as  so  many  feet 
per  minute  per  second. 

*NOTE:  It  is  evident  that  just  as  angular  velocity  of  a  point 
expressed  in  radius X radians  =  linear  velocity  of  the  point,  so  also: 

Angular  acceleration  in  radians / sec2  Xradius  in  ft.  =  linear  accel- 
eration in  ft/ sec2. 


ELEMENTARY  PRACTICAL  MECHANICS  115 

8.  If  the  initial  velocity  of  a  body  is  1000  ft/sec  due 
north,  and  its  acceleration   is  50  ft /sec2  due   south,  find 
the  velocity  of  the  body  at  the  end  of  25  seconds. 

9.  A  ball  is  shot  vertically  upward  with  a  velocity  of 
640  ft /sec,  and  the  acceleration  of  gravity  is  32  ft /sec2. 
How  long   before   the   velocity   of  the   body   will   be    (a) 
512    ft /sec    upward?       (b)    Zero?      (c)    96    ft /sec    down- 
ward?    (d)  640  ft /sec  downward? 

10.  The  velocity  of  a  body  has  changed  from  8  ft  /sec 
due  east  to  212  ft/sec  due  west  in  6  seconds.     What  has 
been  its  acceleration? 

11.  A  wheel  making  120  revolutions  per  minute  has  its 
velocity  increased  in   10  seconds  to   180   revolutions  per 
minute.     Find    the    acceleration     (a)   in    revolutions    per 
minute  gained  each  second;    (6)  in  radians  per  second  per 
second. 

12.  A  point   on  a  rotating   body  is    moving  50  ft /sec. 
The  angular  velocity  of  the  body  is  increased  1  revolution 
per  minute  each  second  for  10  seconds.     What  will  then 
be  the  velocity  of  the  point  in  ft /sec  if  it  is  situated  5  ft. 
from  the  axis  of  rotation? 

66.  Average  Velocity  and  Distance. — Suppose  a  body 
to  move  with  a  uniform  acceleration  and  let  its  velocity 
at  the  time  we  begin  to  observe  its  motion  (initial  velocity) 
be  6  ft/sec.  At  the  end  of  5  seconds  suppose  the  velo- 
city to  have  become  26  ft/sec.  Then, 

Change  of  velocity  26-6  =  20  ft/sec. 

26-6     20 

Acceleration  = = — =4  ft /sec  in  one  sec. 

5          5 

The  average  velocity  during  the  5  seconds  will  be  one- 
half  the  sum  of  the  initial  and  the  final  velocities,  or 

A  _i_  o£» 

Average  velocity  =  — - —  =  16  ft/sec. 


116  ELEMENTARY  PRACTICAL  MECHANICS 

The  total  distance  traversed  will  be  the  average  velocity 
multiplied  by  the  time,  or 

Distance  =  16X5  =  80  ft. 

It  is  of  the  utmost  importance  for  the  student  to  master 
thoroughly  these  simple  relations.  They  are  stated  for- 
mally as  follows,  and  apply  equally  to  rotation  and  trans- 
lation: 

Change  of  velocity  =  Acceleration  X  time; 
Final  velocity  =  Initial  velocity  +  change  of  velocity; 
Average  velocity  =  J  (initial  velocity  +  final  velocity); 
Distance  =  Average  velocity  X  time. 

NOTE:  It  should  be  noticed  that  these  statements  are  perfectly 
general  for  all  cases  of  uniformly  accelerated  motion.  Thus,  if  a 
body  start  from  rest,  its  initial  velocity  is  0.  Also,  if  the  initial 
velocity  is  in  one  direction  and  the  acceleration  is  in  the  opposite 
direction,  the  rules  above  still  apply  if  we  give  to  one  direction 
the  +  sign  and  to  the  other  the  —  sign. 

ILLUSTRATION.- — A  body  is  moving  N.  with  a  velocity 
of  40,  ft/  sec,  and  a  force  acts  S.  so  as  to  give  it  an 
acceleration  S  of  8  ft/sec2.  What  will  be  its  velocity 
at ,  the  end  of  7  seconds,  and  how  far  from  the  start  will 
it  be  at  the  end  of  that  time?  Call  the  initial  velocity  +40 
and  the  acceleration  -8.  Then 

Change  of  velocity  =  -8X7=  -56; 
Final  velocity         =  +  40  -  56  =  - 16; 

40-16 

Average  velocity    = •  —  12; 

A 

Distance  12x7=84  ft. 


ELEMENTARY  PRACTICAL  MECHANICS  117 

That  is,  the  body  will,  at  the  end  of  the  7  seconds,  be 
84  ft.  N.  of  its  original  position,  and  will  be  moving  S. 
with  a  velocity  of  16  ft/ sec. 

(What  has  actually  happened  if  the  full  history  of  the 
motion  is  required,  is  that  the  body  has  moved  N.  until 
brought  to  rest,  then  has  started  S.  from  zero  velocity 
and  moved  until  its  velocity  is  16  ft/ sec  S. 

40 
To  lose  all  velocity  K  will  require  —  =  5  seconds,  and 

8 

the  distance  moved  N.  will  be  X5  =  100  ft. 

2i 

The  time  it  moved  south  is  therefore  7  —  5  =  2  seconds. 
Velocity  at  the  end  of  the  7th  second  is  therefore  2X8  =  16 

0  +  16 

ft/sec    S.     Distance    moved   S.    after   stopping  = X2 

2i 

=  16  ft.     Therefore  the  body  is  100-16  =  84  ft.  N.  of  start- 
ing point.) 

67.  Equations  for  Accelerated  Motion. — To  express  the 
relations  of  the  various  items  of  velocity,  time,  distance, 
etc.,  in  accelerated  motion  by  simple  equations,  we  may 
make  use  of  the  following  notations: 

u  =  initial  velocity; 

v  =  final  velocity; 

t  =  time; 

a  =  acceleration; 

S  =  distance. 

Then  a  t  =  change  in  velocity, 

and  v  =  u  +  at. 

u+v 
Also,         Average  vel.  =  — —     or 

£  i 

and  Distance  =  (u  +  \a  t)  X  t  =  u  t  +  £a  t2. 


118  ELEMENTARY  PRACTICAL  MECHANICS 

68,  Accelerated  Motion  from  Rest. — If  a  body  starts 
from  rest,  that  is,  if  w  =  0,  u  disappears  from  the  preceding 
equations  and  S  =  ^a  t2. 

The  distance  passed  over  by  a  body  starting  from  rest 
and  moving  with  a  uniform  acceleration,  therefore  varies 
as  the  SQUARE  OF  THE  TIME.  Hence,  if  the  body  moves 


FIG.  116. — Apparatus  for  Study  of  Accelerated  Rotary  Motion. 

10  ft.  the  first  second,  it  will  move  2X2X10  =  40  ft.  in 
2  seconds,  3X3X10  =  90  ft.  in  3  seconds,  etc. 

The  student  should  test  the  preceding  statements  for 
motion  of  translation  by  use  of  the  At  wood's  machine, 
body  moving  down  an  incline,  or  any  of  the  well  known 
apparatus  for  studying  the  laws  of  accelerated  motion. 

Fig.  116  shows  a  convenient  device  for  studying  acceler- 
ated motion  of  rotation.  The  moving  body  consists  of 
a  pulley  Pf  accurately  turned  and  centered,  mounted 


ELEMENTARY  PRACTICAL  MECHANICS  119 

upon  hardened  cone  bearings.  Motion  is  caused  by  a 
weight  W  descending  and  unwinding  a  cord  from  the 
drum  C.  To  obtain  a  time  record  a  strip  of  paper,  such 
as  is  supplied  in  rolls  for  adding  machines,  is  placed  tightly 
around  the  pulley  and  the  overlapping  ends  pasted  together. 
The  paper  is  then  smoked  until  evenly  coated  with  soot. 
As  the  pulley  is  rotated,  the  vibrations  of  an  electrically- 
driven  tuning  fork  are  traced  upon  the  paper. 


Section  at  Beginning  of  Trace. 


Section  of  Middle  of  Trace. 
FIG.  117. — Specimen  Tracing  of  Tuning  Fork. 

In  use,  the  fork  is  first  placed  on  the  paper  and  by  a 
transverse  motion  a  straight  line  is  drawn  to  mark  the 
zero  from  which  to  measure  distances.  The  wheel  is  then 
released  and  the  trace  taken  for  about  two  revolutions, 
when  the  fork  is  raised  and  the  battery  switch  opened. 
Several  records  may  be  obtained,  side  by  side,  by  shifting 
the  fork.  The  paper  is  then  cut  arid  fixed  by  passing 
through  thin  shellac.  A  specimen  tracing  is  shown  in 
Fig.  117. 

The  first  two  or  three  waves  of  the  tracing  are  not 
legible  as  all  the  soot  is  removed.  To  locate  the  10th 


120 


ELEMENTARY  PRACTICAL  MECHANICS 


wave,  a  preliminary  determination  of  the  acceleration  fo£ 
a  10-wave  period  is  made;  the  10-wave  period  is  then 
used  throughout  as  the  unit  of  time.  In  this  preliminary 
determination  an  arbitrary  point  of  reference,  R,  a  couple 
of  centimeters  from  the  starting  line  of  the  trace,  is  chosen, 
and  the  distances  passed  over  in  successive  periods  of 
10  waves  is  measured  from  this  point.  The  gain  in  velocity 
per  10  waves,  in  a  10-wave  period  is  found  then  by  finding 
the  gain  in  distance  of  each  10-wave  period  over  the  pre- 
ceding. The  accompanying  table  shows  the  data  of  such 
a  preliminary  determination  of  acceleration. 


Distance  from  R. 

Distance  in 
10  Waves. 

Distance  Gained 
in  10  Waves. 

First  10  waves 

3  99 

6.11 

2  09 

Second  10  waves  

10  10 

8  20 

2  15 

Third  10  waves    

.   18.30 

10.35 

2  05 

Fourth  10  waves  

.   28.65 

12.40 

2.10 

Fifth  10  waves             .    ... 

.  41  05 

14  50 

2  10 

Sixth  10  waves  

.   55.55 

16.50 

2.10 

Seventh  10  waves  .  .    . 

72  15 

18  70 

Eighth  10  waves  

.  90.85 

From  the  table  it  is  seen  that  the  average  gain  in  velocity 
per  unit  time  (10  waves),  which  takes  place  in  each  10 
waves  period,  is  2.10  cms.  The  gain  in  speed  during  the 
first  unit  of  time  when  the  trace  is  illegible  is  therefore 
also  2.10  cms.;  and  as  the  initial  velocity  was  zero,  the 

0  +  2.10 

average  velocity  must  have  been  ,  or  1.05  cm.  per 

& 

unit  time.  This  distance  may  now  be  measured  off  from 
the  starting  point  and  marked  as  the  first  10-wave  point. 
By  counting,  the  succeeding  10-wave  distances  are  marked. 
Measurements  are  then  continued  from  the  10-wave  point 


ELEMENTARY  PRACTICAL  MECHANICS 


121 


thus  fixed,  giving  the  distances  passed  over  in  the  actual 
time  elapsed  since  the  release  of  the  wheel  or  true  zero 
of  time.  From  this  record  of  time  and  distance  curves 
may  be  plotted  as  in  Fig. 
118,  verifying  the  laws  of 
accelerated  motion  of  rota- 
tion, i.e., 


and 


Velocity  octime 
Distance  oc  (time)2. 


y 

80  32 

^ 

/ 

70  28 

2 

* 

_60  24 

3 

y 

n 

I 

3 

£ 

0  60  .2  20 

\ 

A' 

i 

o 

i 

^ 

k 

S  40  1  16 

3 

j 

yc 

»v 

1   " 

f 

7 

< 

30  12 

/<* 

./ 

!• 

/ 

f 

/< 

•>v 

2 

/ 

/ 

10   4 

/ 

i 

/ 

n   A 

/ 

100 

"Quits  of  Time  (Waves) 

2000    4000    6000    8000    10000 

Units  of  Time2(Wayes) 

FIG.  118. 


Since  the  distances  on  the 
rim  of  the  wheel  are  pro- 
portional to  the  angular  dis- 
placement corresponding,  the 
laws  of  angular  acceleration 
under  a  constant  torque,  are 
seen  to  be  exactly  analogous 
to  the  laws  of  accelerated  mo- 
tion in  a  straight  line  under  the  action  of  a  constant  force. 

69.  Motion  of  a  Body  under  the  Influence  of  Gravity. 
A  body  falling  under  the  influence  of  gravity,  if  we  neg- 
lect the  resistance  offered  by  the  air,  furnishes  a  common 
instance  of  uniformly  accelerated  motion. 

In  this  course,  we  shall  always  assume  the  acceleration 
produced  by  gravity  on  all  bodies  to  be 


or 


32  ft.  per  sec2, 
980  cm.  per  sec2. 


This  value  is  represented  by  the  symbol  g.    The  equa- 
tions for  falling  bodies  are  therefore, 


122  ELEMENTARY  PRACTICAL  MECHANICS 

(1)  S=ut  +  %gt2  for  bodies  having  initial  velocity. 

(2)  $  =  %gt2  for  bodies  falling  freely  from  rest. 


Since  v=gt    and    t=—  , 

9 


Substituting  in  (2)  we  have  v  =  v2gS  as  the  expression 
for  the  velocity  acquired  by  a  body  in  falling  freely  from 
rest  through  the  distance  S. 

The  value  of  the  force  of  gravity  and  therefore  of  the 
acceleration  that  it  produces,  g,  varies  somewhat  from 
place  to  place  on  the  earth's  surface,  being,  in  general, 
least  at  the  equator  and  increasing  towards  the  poles. 
It  is  also  greater  at  the  surface  of  the  earth  than  at  points 
below  or  above  the  surface.  These  variations,  however, 
are  small  and  for  most  purposes  the  values  32  ft/sec2 
and  980  cm/sec2  are  sufficiently  exact  for  our  use. 

70.  Solution  for  a  Body  Projected  Vertically  Upward  or 
Downward.  —  Suppose  a  ball  is  projected  vertically  upward 
with  a  velocity  of  192  ft  /sec.  Find:  (a)  The  time 
during  which  it  will  rise;  (6)  the  height  to  which  it 
wrill  rise;  (c)  its  velocity  after  8  seconds;  (d)  its  velocity 
on  striking  the  ground. 

(a)  Initial  velocity  =  —  192, 

Acceleration      =       32. 

Therefore,  since  its  velocity  at  the  highest  point  of  its 
journey  (final  velocity)  is  0,  we  have 


t=&  seconds. 


ELEMENTARY  PRACTICAL  MECHANICS  123 

(6)  Initial  velocity  =  -192; 
Final  velocity  =  0; 
Aver,  velocity  =  —  96; 

Distance  =  -  96x6  =-576   ft.,   or  body  will 

rise  576  ft. 

(c)  Change  of  velocity  in  8  sec.  =  32  X  8  =  256  ; 

Final  velocity  =-192  +  256  =+64   ft  /sec.     Or  body 
will  be  moving  64  ft  /sec  downward. 

(d)  It  took  6  seconds  to  reach  the  highest  point  where 
the    velocity    was    0.     Considering   now    only   the   return 
journey,  we  may  call  the 

Initial  velocity  =  0; 
Acceleration      =32; 

Final  velocity  =32£  (where  t  is  the  time  of  return 
journey.) 

Aver,  velocity  =16£; 
Distance  =576  ft.; 

or  576  = 


and  the  velocity  on  reaching  the  ground  will  be  6X32  =  192 
ft/sec. 

Thus  we  see  that,  in  this  case,  the  time  of  rise  will  be 
equal  to  the  time  of  the  return.  The  whole  time  of  flight 
is  12  seconds.  Also  the  downward  velocity  on  reaching 
the  ground  will  be  numerically  equal  to  the  upward  velocity 
with  which  the  body  started. 


124  ELEMENTARY  PRACTICAL  MECHANICS 

In  the  case  of  a  body  projected  vertically  downward  we 
may  apply  the  laws  of  accelerated  motion  in  the  same 
way,  the  initial  velocity  and  the  acceleration  of  gravity 
being  here  in  the  same  direction. 

PROBLEMS 

1.  A  sled  moves  from  rest  until  in  12  seconds  it  has 
acquired  a  velocity  of  60  ft.  a  second.     Find  the  accelera- 
tion, the  average  velocity,  the  velocity  at  the  end  of  7 
seconds,  and  the  distance  traversed  in  the  12  seconds. 

2.  If  an  engine  takes  24  seconds  to  get  up  a  velocity 
of  300   revolutions  per  minute  in  its   fly-wheel,   find    (a) 
the  angular  acceleration,    (6)    the  angular  velocity  at  the 
end  of  15  seconds,  (c)  the  average  angular  velocity  during 
the  process  of  speeding  up. 

3.  The  acceleration  of  a  body  starting  from  rest  is  3 
ft/sec2.     Plotting    times    and    velocities    for    the    end    of 
each  of  the  first  5  seconds,  get  the  time-velocity  diagram 
and  draw  on  it  the  line  representing  the  average  velocity. 
Determine  also  the  distance.     Plot  time  as  abscissae. 

4.  For  6  seconds  after   starting,  a  body  has  a  positive 
acceleration    of    3  ft/sec2.     It    then    proceeds    with    the 
velocity  already  acquired.     Draw  the  time-velocity  diagram 
with  the  final  velocities  for  the  first  10  seconds.     (Time  as 
abscissae.) 

5.  A  bicycle  and  rider  start  from  rest  at  the  top  of  a 
hill.     At  end  of  10  seconds  the  foot  of  the  hill  is  reached 
with  a  speed  of  3  feet  per  second.     Find  the  acceleration 
(considered  as  uniform)  of  the  wheel  during  the  descent. 

6.  A  body  going  with  a  uniform  speed  of  50  ft.  per  second 
is  acted  upon  by  a  force  which  gives  it  an  acceleration 
per  second  of  5  ft.  per  second.     How  fast  will  the  body 
be  going  at  the  end  of  30  seconds? 

7.  A  body  is  going  28  ft.  per  second,  when  its  velocity  is 
increased  4  ft.  per  second  each  second  for  6  seconds.     Find 
the  velocity  at  the  end  and  the  distance  traveled. 


ELEMENTARY  PRACTICAL  MECHANICS  125 

8.  The   velocity    of   a   body    changes   from   400  ft.    per 
second  to  180  ft.  per  second  in  5  seconds.     Find  the  ac- 
celeration and  the  distance.  * 

9.  An  acceleration  of  30  (feet  and  seconds)  equals  what 
acceleration  in  feet  and  minutes? 

10.  A  railroad  train  starts  from  rest  at  a  station  and 
increases  its  speed  uniformly  at  the  rate  of  J  ft:  per  second 
each  second.     How  fast  will  it  be  going  at  the  end  of  20 
seconds? 

11.  Express  the  speed  of  Problem  10  in  terms  of  miles 
per  hour. 

12.  A  body  going  at  the  rate  of  500  ft.  per  second  has  its 
speed  uniformly  accelerated  until  at  the  end  of  10  minutes 
its  speed  is   1000  ft.   per  second.     Find  the  acceleration 
per  minute  in  feet  per  second. 

13.  A  body  going   100  ft.  per  second  has  its  velocity 
decreased  to  \  ft.  per  second  in  going  100  ft.     Find  the 
acceleration. 

14.  A  train  running  at  rate  of  30  miles  per  hour  is  stopped 
by   a   sudden   application   of   the   brake.     What   negative 
acceleration  must  be  given  in  order  that  the  train  may 
be  brought  to  rest  (at  a  uniform  rate)  in  20  seconds? 

15.  A  wheel  starting  from  rest  has  its  speed  increased 
uniformly  each  second  15  degrees   per  second.     What  will 
be  its  velocity  of  rotation  in  degrees  per  second  at  the 
end  of  one-half  minute? 

16.  If  the  wheel  in  the  preceding  problem  were  already 
making  10  revolutions  per  second  when  the  increase  began, 
what  would  be  its  velocity  at  the  end  of  one-half  minute? 

17.  A  car  is  moving  uniformly  at  the  rate  of  3  ft.  per 
second.     What  uniform  acceleration  must  be  given  it  so 
that  at  end  of  10  seconds  it  may  be  going  at  speed  of  30  ft. 
per  second? 

18.  An  elevator  starting  from  rest  moves  with  an  accelera- 
tion of  1  ft /sec2  for  ten  seconds.    Find  final  velocity,  average 
velocity,  and  distance. 


126  ELEMENTARY  PRACTICAL  MECHANICS 

19.  A  railroad  train,  going  30  miles  per  hour,  increases 
its  speed  uniformly  each  second,  ^  ft.  per  second  for  5 
minutes.     At  what  rate  will  it  then  be  moving  (in  miles 
per  hour)   and  how  far  will  it  have  gone  during  the   5 
minutes? 

20.  A  moving  body  has  its  velocity  changed  uniformly 
from  1500  ft.  per  second  to  300  ft.  per  second  in  30  seconds. 
Find  the  negative  acceleration.     How  far  will  it  go  mean- 
time? 

21.  A  car  sliding  down  a  grade  changes  its  velocity  from 
20  ft.  per  second  to  4  ft.  per  second.     If  this  is  done  with 
a  uniform  negative  acceleration  per  second  of  2  ft.  per 
second,  what  time  elapses  during  the  change?     How  far 
does  the  car  go  meanwhile? 

22.  Find  the  acceleration  of  a  body  which  starting  from 
rest  acquires  a  velocity  of  300  cms.  per  second  while  moving 
through  a  distance  of  1200  centimeters. 

23.  A  body  starting  from  rest  increases  its  speed  each 
second  5  ft.  per  second.     How  far  will  it  move  during  the 
fourth  second? 

24.  A  wheel  making  5  revolutions  per  second  has  its 
velocity  of  rotation  increased  each  second  30  degrees  per 
second.     How    many    revolutions    per    second    will    it    be 
making  at  the  end  of  10  seconds? 

25.  A  street  car  has  its  speed  diminished  from  40  ft.  per 
second  to  15  ft.  per  second.     If  the  brakes  were  applied 
with   such   force   as   to   diminish   the   car's   velocity   each 
second  5  ft.  per  second,  how  far  did  the  car  go  in  the 
meantime? 

26.  To  what  height  and  for  what  time  will  a  body  rise 
if  projected  vertically  upward  with  a  velocity  of  100  ft.  a 
second? 

27.  With  what  initial  velocity  must  a  body  start  in  order 
to  rise  vertically  700  ft.  before  losing  all  its  velocity? 

28.  What  initial  velocity  must  a  body  have  if  it  is  to 
move    with    a    positive    acceleration    of    4  ft/sec2    for    13 
seconds,  and  pass  over  500  ft.? 


ELEMENTARY  PRACTICAL  MECHANICS  127 

29.  A  body  going  150  ft.  a  second  has  its  speed  decreased 
to  50  ft.  per  second  in  going  240  ft.     Find  the  accelera- 
tion. 

30.  A  ball  is  thrown  vertically  downward  from  the  top 
of  a  tower  550  ft.  high.     In  5  seconds  it  is  25  ft.  from  the 
ground.     What  was  its  initial  velocity  and  when  will  it 
reach  the  ground? 

31.  A  ball  is  dropped  from  a  certain  point  and  \  of  a 
second  afterwards  another  ball  is  dropped  from  the  same 
point.     How  far  apart  will  they  be  when  the  second  ball 
has  fallen  2  seconds? 

32.  A  balloon  is  rising  at  the  rate  of  20  ft.  a  second 
when  a  sand  bag  is  dropped  from  it.     In  what  time  and 
with    what    velocity    will    the    bag    reach    the    ground    if 
the    balloon    was    half    a    mile    high    when   the    bag    was 
dropped? 

33.  A  stone  thrown  vertically  upward  from  the  ground 
under  an  arch  is  seen  to  just  touch  the  arch  and  return 
to  the  ground  in  4.5  seconds  after  it  started.     How  high 
is  the  arch? 

34.  Find  the  acceleration  if  a  train  starting  with  a  ve- 
locity   of    15    miles    per   hour   passes    over   350  ft.    in    10 
seconds. 

35.  With  what  velocty  must  a  body  be  projected  ver- 
tically upward  to  attain  a  height  of  6000  ft.? 

36.  A  wheel  starting  from  rest  is  given  an  acceleration 
each    second    of   3.4    radians   per   second.     Through   what 
distance  will  a  point  2  ft.  from  the  axis  of  rotation  move 
in  20  seconds? 

37.  The  velocity  of  a  pair  of  rollers  in  a  paper  mill  changes 
from   80   revolutions   per  minute   to    120   revolutions   per 
minute,  in  30  seconds.     If  the  paper  is  passed  between 
them  without  slippage,  what  length  of  paper  will  be  passed 
in  30  seconds,  assuming  the  diameter  of  the  roll  as  20 
inches? 


128 


ELEMENTARY   PRACTICAL   MECHANICS 


71.  Path  of  Bodies  Projected  Horizontally  or  at  an  Angle 
of  Elevation  less  than  90°. — 1.  Motion  of  a  body  projected 
from  a  height  with  a  uniform  horizontal  velocity.  If  a  body 
is  projected  from  a  height  in  a  horizontal  direction,  the 
velocity  of  projection  and  the  acceleration  of  gravity 
are  at  right  angles  to  each  other  and  the  body  will. move 
in  a  curve.  The  position  of  the  body  at  any  time,  if  we 
disregard  air  resistance,  may  be  determined  by  remember- 
ing that  the  horizontal  velocity  will  in  no  way  affect  the 
rate  of  vertical  fall  under  the  influence  of  gravity,  nor 
will  the  fact  that  the  body  is  falling  in  any  way  affect  the 

uniform  horizontal  veloc- 

Y     40Q/    A'    400'    B'    400'    c'    400'   x'      ity  with  which  it  started. 

This  is  really  an  illus- 
tration of  the  truth  of 
"  Newton's  Second  Law 
25G/  of  Motion."  Thus,  if  a 
body  be  projected  hori- 
zontally from  the  top  of 
a  tower  with  a  velocity 
of  400  ft.  a  second,  it 
will  move  in  3  g&GOUdfi 

a  distance  of  3X400  =  1200  ft.  out  from  the  tower.  IB 
the  same  time  it  will  of  course  have  fallen  with  an  accel- 
eration of  32  ft /sec2. 

Its  average  vertical  velocity  will  be  i(32X3)=4§  £& 
per  second,  and  the  vertical  distance  fallen  48X3™  144  fe. 
Then  at  the  end  of  the  3  seconds,  if  we  neglect  modifica- 
tions of  its  motion  due  to  the  air  through  which  it  !l£E 
passed,  it  will  be  144  ft.  below  the  top  of  the  tower  aaji 
1200  ft.  out  from  it. 

If  the  positions  of  a  body  under  the  conditions  just 
described  are  calculated  for  several  intervals  of  time  and 
the  results  plotted,  a  curve,  as  in  Fig.  119,  will  be  obtained 


FIG.  119. 


ELEMENTARY  PRACTICAL  MECHANICS 


129 


showing  the  position  of  the  body  at  every  instant,  and 
the  general  shape  of  its  path.  This  curve  is  a  parabola. 

2.  Motion  of  a  body  starting  with  a  uniform  velocity  in  a 
direction  inclined  to  the  horizontal  at  less  than  90  degrees. 
Suppose  a  body  projected  from  0  (Fig.  120)  along  the  line 
OE,  making  an  angle  d  with  the  horizontal.  Let  OA 
represent  the  velocity  with  which  the  body  is  projected 
from  0.  Then  if  gravity  did  not  act,  the  body  in  one 
second  would  be  at  A,  in 
2  seconds  at  B,  in  3  sec- 
onds at  (7,  in  4  seconds  at 
D,  etc.,  where  OA=AB  = 
BC.  But  since  gravity 
also  acts,  the  body  will  be 
below  A  Sit  the  end  of  one 
second,  by  an  amount  equal 
to  the  distance  it  will  fall 
in  one  second,  or  it  will  be  v 
16  ft.  vertically  below  A. 
Similarly,  at  the  end  of  2 
seconds  it  will  be  64  ft. 
vertically  below  B,  etc.  In 
this  way  we  may  construct  a  curve  giving  the  path  of  the 
body  when  we  have  given  the  amount  and  the  angle  of  the 
original  velocity. 

The  greatest  height,  S,  and  "  range/'  OR,  of  this  curve 
may  be  computed  as  follows: 

Resolve  OA  into  its  component  horizontal  and  vertical 
velocities,  H  and  V. 

H  =  0 A  cos  d    and     V  =  0 A  sin  d, 

from  which  the  numerical  values  of  H  and  V  may  be 
calculated  if  OA  is  given.  Now  the  horizontal  and  the 
vertical  motions  may  be  considered  separately.  The 


FIG.  120. 


130  ELEMENTARY  PRACTICAL  MECHANICS 

component  V  is  the  initial  vertical  velocity  and  H  is  the 

V 

uniform  horizontal  velocity.     Thus,  —  =  t,  the  time  during 

32 

which  body  will  continue  to  rise.     Therefore, 


Since  the  body  will  require  the  same  time  to  fall  as 
was  required  in  rising,  the  whole  time  of  flight  will  be  2t. 

Therefore  "  horizontal  range/7  OR  =  Hx2t.  No  allow- 
ance is  here  made  for  air  resistance,  wind,  etc.  These 
effects  are  variable  with  the  shape  of  the  projected  body, 
initial  velocity,  etc.,  and  special  corrections  must  be  applied 
for  any  actual  case  of  projection,  as,  for  example,  in  the 
firing  of  long-range  guns. 

PROBLEMS 

1.  A  stone  is  thrown  horizontally  out  from  the  top  of  a 
cliff   248  ft.    high    with    a   velocity    of    70  ft.    per   second. 
Draw  a  diagram  to  scale  showing  starting  point,  point  of 
fall,  and  four  intermediate  positions. 

2.  When  and  where  will  a  ball  strike  the  ground  if  it 
is  thrown  horizontally  out  from  the  top  of  a  tower  200  ft. 
high  with  a  velocity  of  20  ft.  per  second? 

3.  A  body  has  simultaneously  a  uniform  velocity  due 
north  of  40  ft.  per  second,  and  an  acceleration  due  east 
of  20  ft /sec2.      Find  its  position  at  the  end  of  the  first, 
second,  third,  fourth,  and  fifth  seconds. 

4.  Plot  the  path  of  the  body  in  Problem  3,  and  from  the 
curve  take  the  position  of  the  body  at  the    end  of  3.6 
seconds. 

5.  A  projectile  is  fired  at  an  angle  of  30  degrees  to>  the 
horizontal  with  a  velocity  of   144  ft.   per  second.     Draw 
a  diagram  of  its  path  showing  its  position  at  the  end  of  each 
of  the  first  5  seconds. 


ELEMENTARY  PRACTICAL  MECHANICS 


131 


6.  Find  the  greatest  height  reached  and  the  horizontal 
range  of  a  projectile  the  velocity  of  which  is  450  ft.  per 
second  at  an  angle  of  60  degrees  to  the  horizontal. 

7.  A   projectile   is   thrown    with   an   initial   velocity   of 
600  cm.  per  second  at  an  angle  of  50  degrees  with  the  hori- 
zontal.   Find  total  time  of  flight  and  the  horizontal  range. 

72.  Simple  Harmonic  Motion. — A  third  type  of  motion, 
as  distinguished  from  uniform  motion  and  uniformly 
accelerated  motion,  is  illustrated  in  the  periodic  swinging 
back  and  forth  between  two  extreme  positions  of  a  pendu- 
lum-bob, of  the  free  end  of  a  flexible  rod  when  clamped 
at  one  end  and  then  pulled  aside  and  suddenly  released, 
of  a  weight  suspended  from  a  spring,  etc.  Motion  of  this 
sort  is  known  as  simple  harmonic 
motion.  All  vibrations  due  to  the 
elastic  properties  of  bodies  are 
simple  harmonic.  The  to-and-fro 
motion  of  the  cross-head  of  a  steam- 
engine  is  approximately  simple  har-  B  [ 
monic. 

The  general  characteristics  of 
simple  harmonic  motion  are  shown 
in  the  following  illustration: 

Imagine  a  point  Pf  to  move  with 
a  uniform  velocity  v  in  a  circular  path  (Fig.  121),  and  imagine 
a  perpendicular  dropped  from  each  successive  position  of 
this  point  to  a  diameter  AB  of  the  circular  path.  As  Pf 
moves  around  the  circle,  point  P,  the  intersection  of  this 
perpendicular  with  the  diameter  will  move  periodically 
back  and  forth  between  the  two  extreme  positions  A  and 
B.  The  time  required  for  P  to  move  from  A  to  B  and 
then  back  again  from  B  to  A,  is  called  the  period  of  P's 
oscillation.  It  is  evidently  the  time  required  for  Pr  to 
travel  once  around  its  circular  path. 


FIG.  121. 


132  ELEMENTARY  PRACTICAL  MECHANICS 

Inspection  of  the  figure  shows  also  that  while  Pr  moves 
with  uniform  velocity,  passing  over  the  equal  arcs,  APf \, 
P'\P'2,  etc.,  in  successive  equal  intervals  of  time,  point  P 
travels  in  the  corresponding  time  intervals  over  distances 
APi,  PiPzy  etc.,  which  are  unequal  and  which  increase  as 
P  moves  toward  the  center  0,  and  decrease  as  P  moves  away 
from  the  center.  The  velocity  of  P  along  the  diameter 
at  its  extreme  positions  A  and  B,  when  Pf  is  moving  for 
an  instant  at  right  angles  to  the  diameter,  is  zero.  At  P'& 
and  the  corresponding  position  on  the  other  semicircum- 
ference,  Pr  is  moving  for  an  instant  parallel  to  the  diameter 
A  B,  hence  at  these  points,  P,  which  is  passing  through  0, 
moves  with  a  maximum  velocity  which  is  equal  to  the 
velocity  of  P',  or  v.  Starting  from  an  extreme  position  as 
A,  therefore,  with  a  velocity  of  zero,  P  moves  with  increas- 
ing velocity  along  its  path  to  0  where  its  velocity  has  a 
maximum  value  v.  From  0  to  B,  the  velocity  of  Pdecreases, 
becoming  zero  when  B  is  reached.  Similar  changes  in 
velocity  take  place  as  P  moves  back  over  its  path  from 
B  to  A.  The  motion  of  P  is  therefore  accelerated,  the 
acceleration  being  always  toward  the  center. 

It  may  be  shown  also  that  the  value  of  P's  acceleration  is 
not  uniform,  but  is  dependent  upon  the  position  of  P  on 
the  diameter  AB.  An  approximate  demonstration  of  this 
important  idea  may  be  made  graphically  as  follows:  Draw 
Fig.  121  to  a  large  scale  and  measure  the  distances  AP\, 
PiP2,  etc.  The  data  given  in  the  table  opposite  was 
obtained  in  this  way. 

If  we  assume  the  time  required  for  P  to  move  over  each 
space  as  APi,  PiP2,  etc.,  as  "unit "  time,  since  this  time 
is  very  short,  we  may  regard  the  distances  APi,  PiP2,  etc., 
measured  along  the  diameter  as  representing  the  average 
velocities  for  each  successive  unit  of  time.  The  values  in 
column  one  of  the  table  were  obtained  in  this  way.  The 


ELEMENTARY  PRACTICAL  MECHANICS 


133 


distance  PiP2— the  distance  APi  therefore  represents  the 
gain  in  velocity  in  unit  time  or  in  other  words,  equals  the 
acceleration  at  PI.  Similarly,  distance  P2Ps  —distance  P\P% 
represents  the  acceleration  at  P2,  etc.  In  this  way  values 
were  found  for  the  accelerations  given  in  column  two. 
The  "  displacement  "  is  the  distance  from  the  center,  0, 
to  the  point  under  consideration.  Thus  the  displacement 
at  PI  is  represented  by  the  distance  OPi,  at  P2  by  the 
distance  OP2,  etc. 


Average  Velocity. 

Acceleration. 

Displacement. 

Displacement 

For  Space. 

Value. 

At. 

Value. 

At. 

Value. 

Acceleration 

AP, 

.60 

1.72 

fl 

1.12 

PI 

11.39 

10.2  . 

P2PS 

2.64 

P2 

.92 

P2 

9.64 

10.4 

P-P* 

3.33 

P3 

.69 

PS 

7.03 

10.2 

Pfl 

3.70 

P4 

.37 

P, 

3.70 

10.0 

OP. 

3  70 

o 

0 

v*  6 

3.33 

P, 

.37 

3.70 

10.0 

PPs 

2.64 

PI 

.69 

PI 

7.03 

10.2 

PP 

1.72 

PS 

.92 

PS 

9.64 

10.4 

P  B 

.60 

P* 

1.12 

P* 

11.39 

10.2 

From  the  values  in  the  table  it  will  be  seen  that  the 
velocity  of  P  increases  from  A  to  0  and  then  decreases  from 
0  to  B]  and  that  the  acceleration  is  zero  at  0  and  a  maxi- 
mum at  points  A  and  B,  also  that  the  direction  of  the 
acceleration  is  always  toward  the  center  0.  The  values  for 
the  fraction  displacement  over  acceleration  given  in  the  last 
column  of  the  table  are  practically  constant,  thus  showing 
that  acceleration  is  proportional  to  displacement. 

And,  finally,  if  we  substitute  the  average  value  for 
displacement  over  acceleration  in  the  equation,  * 


m_<2  ^/displacement 

»    Q  r»r>CkloT»a  4- 1 r\-r»    ' 


acceleration  J 


134  ELEMENTARY  PRACTICAL  MECHANICS 

and  solve  for  T,  we  have  the  approximate  value  20,  which 
we  see,  since  each  space  represents  unit  time,  is  the  time 
required  for  a  complete  oscillation  of  P  along  the  diameter. 
A  body  having  a  motion  similar  to  that  of  point  P  in 
the  preceding  illustration  is  said  to  have  simple  harmonic 
motion.  The  characteristics  of  simple  harmonic  motion  are: 

1.  It  is  in  the  nature  of  a  periodic  oscillation  back  and 
forth  between  two  extreme  positions. 

2.  The  velocity  of  the  moving  body  is  zero  at  its  posi- 
tions of  greatest  displacement  from  the  center  of  its  path, 
and  a  maximum  when  passing  in  either  direction  through 
the  center. 

3.  The  body  is  constantly  accelerated  toward  the  center 
of  its  path. 

4.  The  acceleration  is  not  uniform  but  is  dependent  upon 
the  position  of  the  body  in  its  path;   the  law  of  variation 
being:   "  Acceleration  is  proportional  to  the  displacement. " 

5.  The  period  of  a  simple  harmonic  motion  is  expressed 
by  the  equation, 


displacement 
acceleration 

73.  Other  Applications  of  Laws  of  Simple  Harmonic 
Motion. — Many  quantities  which  undergo  periodic  changes 
in  value  do  so  in  accordance  with  general  laws  similar  to 
those  of  simple  harmonic  motion.  An  important  case  of 
this  nature  is  furnished  by  the  variation  of  pressure  at  the 
brushes  of  an  alternating  current  generator.  The  recurring 
cycle  of  values  through  which  such  voltage  passes,  together 
with  its  periodic  reversal  in  direction,  follows  laws  approxi- 
mately identical  with  those  governing  the  velocity  of  the 
point  P  in  the  illustration  just  discussed.  The.  rate  of  change 
of  voltage  is  not  uniform  but  undergoes  changes  similar  to 


ELEMENTARY  PRACTICAL  MECHANICS 


135 


the  changes  in  the  value  of  the  acceleration  of  the  point  P, 
the  governing  factor  being  the  position  of  the  generating 
coil  in  its  revolution. 

74.  Phase  of  a  Simple  Harmonic  Motion.-^-Since  both  the 
determining  factors  of  a  motion,  velocity,  and  acceleration 
are  dependent  in  simple  harmonic  motion  upon  the  posi- 
tion of  the  moving  body  in  its  path,  it  is  convenient  to 
have  some  easy  way  of  fixing  the  stage  arrived  at  in  the 
oscillation.     Thus  while  point  P  in  our  illustration,  Fig. 
121,  is  the  body  actually  moving  with  a  harmonic  motion, 
it  is  convenient  to  project  P  back  upon  an  imaginary  circle 
and  fix  its  position  and  therefore   all   conditions   of  the 
harmonic  motion  by  locating  point  P'  on  the  "  circle  of 
reference. "     The  position  of  the  imaginary  radius  vector 
OP',  assumed  to  start  from  the  refer- 
ence line   OA    (see  Fig.   121)    and   to 

rotate  in  the  direction  of  the  arrows, 
then  shows  clearly  the  stage  which  has 
been  reached  in  the  simple  harmonic 
motion.  The  angle  0  between  this 
radius  vector  and  OA,  is  known  as  the 
Phase  of  the  simple  harmonic  motion. 

75.  The  Sine  Curve.— If  the  point  P, 
moving  as  in  Fig.  121,   were    able  to 
make   a  tracing  upon  a  surface  held 
under   the   circle,  and   if  this    surface 
were  moved  uniformly  in  a  direction  at 
right  angles  to  A B,  the  result  would 
be  a   curve,   as  ACDEFG    (Fig.   122). 
This   curve,  which  is   characteristic  of 
simple    harmonic    motion,  is  called   a  F 
Sine  Curve,  since  the  ordinates  of  points 

on  the  curve  are  proportional  to  the  sines  of  the  correspond- 
ing angles  of  phase. 


136  ELEMENTARY  PRACTICAL  MECHANICS 

If  we  imagine  the  abscissae  of  this  curve  to  represent 
time,  or  some  corresponding  quantity,  the  sine  curve 
represents  clearly  the  successive  stages  in  the  simple  har- 
monic motion.  Curve  CDEFG  represents  a  complete 
cycle  of  changes.  Sine  curves  are  commonly  used  in  rep- 
resenting periodic  changes  in  an  alternating  current,  alter- 
nating current  voltages,  etc. 

76.  Pendulums. — A  simple  pendulum  consists  of  a  ball 
suspended  by  a  thread  so  light  and  flexible  that  its  mass 
and  stiffness  may  be  disregarded.  Such  a  pendulum 
furnishes  an  illustration  of  simple  harmonic  motion,  for 
if  set  vibrating,  the  ball  will  pass  periodically  back  and 
forth  over  its  path,  with  a  maximum  velocity  at  the 
middle  point  of  its  swing,  and  zero  velocity  at  the  ends 
where  the  direction  of  motion  reverses. 

Experiment  and  theory  both  show  that  the  period  of  a 
pendulum  (i.e.,  the  time  for  a  complete  oscillation)  is 
independent  of  the  mass  of  the  pendulum-bob  arid  prac- 
tically also  of  the  length  of  arc  through  which  it  swings, 
and  dependent  only  upon  the  length  of  the  pendulum 
and  the  value  of  gravity.  '  The  law  of  the  pendulum  is  ex- 
pressed by  the  equation, 


where  T  is  the  period,  I  the  length  of  the  pendulum,  and 
g  the  acceleration  due  to  gravity  at  the  locality. 

77.  Experimental  Tests  of  the  Laws  of  the  Pendulum. — 
The  relation  of  the  period  of  a  pendulum  to  length  may 
readily  be  shown  experimentally  by  determining  the 
period  of  a  pendulum  consisting  of  a  metal  ball  suspended 
by  means  of  a  silk  thread,  when  the  length  of  the  pendulum 
(measured  from  support  to  center  of  ball)  is  successively 


ELEMENTARY  PRACTICAL  MECHANICS 


137 


1600,  900,  400,  and  100  millimeters  long.  The  ratio  of  the 
square  roots  of  the  lengths  is  then  40:30:20:10  or  4:3:2:1. 
The  corresponding  periods  will  be  found  to  be  practically 
4:3:2:1,  or  the  "  period  of  a  pendulum  varies  as  the 
square  root  of  the  length." 

Fig.  123  shows  a  form  of  pendulum  which  may  be  used 
to  demonstrate  the  relation 
of  period  to  force  acting. 
This  pendulum  may  be  in- 
clined at  any  desired  angle 
to  the  vertical  and  its  period 
then  determined.  Angles 
with  the  vertical  are 
measured  by  means  of  a 
vernier  attachment  upon 
th'e  base  C.  The  bearings 
of  the  pendulum  are  steel 
cones  giving  little  friction. 

When  the  axis  of  the 
pendulum  is  inclined  at  an 
angle  6  with  the  vertical, 
the  force  of  gravity  G  is 
resolved  into  two  compo- 
nents, a  and  6,  Fig.  124. 
One  of  these,  a,  is  counter- 
acted by  the  bearings.  The  FIG.  123.-Inclined  Pendulum  for 
/  fc  proving  relation  of  Period  of  Vib- 

other,  6,  causes  the  pendu-        ration  to  Force  Acting. 

lum  to  oscillate. 

It  is  evident  from  the  figure  that  6  =  (?eos#;  therefore 
by  changing  the  angle,  different  values  of  the  force  causing 
oscillation  may  be  obtained.  Thus,  when  the  pendulum 
is  vertical,  a  =  0,  6  =  G;  at  an  angle  0  =  30°,  6  =  G  cos  30° 
=  .866(7,  etc.  Assuming  (?  =  1,  a  series  of  relative  values 
for  the  component  b  may  be  obtained.  Plotting  these 


138 


ELEMENTARY  PRACTICAL  MECHANICS 


results  with  corresponding  periods  as  ordinates  will  give 

curves  as  A  and  J5,  Fig. 
125.  It  is  evident  from 
the  straight  line  B  that  the 

period  varies  as 

vforce 

This  law,  which  may  also 
be  stated,  "  The  square 
of  the  number  of  vibra- 
tions varies  directly  as 
the  force  acting,"  is  a 
general  law  for  periodic 
vibration.  An  important 
illustration  is  seen  in  the 

case  of  a  magnetic  needle  oscillating  in  a  magnetic  field. 

The    period  of    such  a  pendulum  will  depend  upon  the 


FIG.  124. 


1.60 
1.20 
0.80 

to  10 

1 

3 

A.  Relation  of  Period  to  Force 
B  .   Relation  of  Period  to       1 

X 

)B 

\ 

V    Fo 

rce 

X 

X 

X 

X 

S 

Ss, 

jj 

X 

j 

^ 

^-€U 

—-  —  , 

-^ 

)A 

/ 

x 

0.00 

x" 

X 

<x 

X 

0. 

JO 

0. 

10 

Force 
0.60 

0. 

50 

1. 

)0 

(Sea 

e  Aj 

0.10             0.80             1.^0             1.60             2.00             2.10             2.80 

(Scale  B) 


FIG.  125. 


intensity  of  field  in  which  it  is  placed,  or  n2  oc  H  where  H 
represents  intensity  of  field. 


ELEMENTARY   PRACTICAL  MECHANICS  139 

78.  Determination   of    Acceleration  Due    to   Gravity. — 

The  acceleration  of  a  freely  falling  body  differs  slightly  for 
different  localities.  Various  methods  have  been  used  for 
determining  its  value,  the  most  common  being  some  form 
of  pendulum.  If  a  simple  pendulum  of  known  length  is 
oscillated  and  its  period  found,  the  values  for  length  and 
period  may  be  substituted  in  the  equation  for  the  pendulum 
and  the  expression  solved  for  "g."  Various  special  forms 
of  the  pendulum,  designed  to  give  more  accurate  deter- 
minations of  period  or  of  length,  have  been  devised  for  this 
experiment.  A  description  of  Borda's  and  Kater's  forms 
will  be  found  in  almost  any  laboratory  manual  of  physics. 
A  device  which  has  the  advantage  of  being  simple  and 
easily  manipulated,  and  which  does  not  require  that  the 
speed  of  the  falling  body  be  lessened  by  a  counterweight, 
as  in  the  Atwood's  machine,  or  by  using  only  a  component 
of  the  force  of  gravitation,  as  on  an  inclined  plane,  is  shown 
in  Fig.  126. 

The  pendulum,  by  means  of  which  both  the  time  and 
the  distance  are  determined,  consists  of  an  oak  rod  about 
9  ft.  long  swinging  on  a  knife-edge  at  C.  The  steel  plates 
on  which  the  knife-edge  rests  are  carried  on  a  platform 
standing  on  leveling  screws  M,  by  means  of  which  the 
pendulum  may  be  caused  to  swing  in  a  true  vertical  plane. 
The  upper  end  of  the  pendulum  is  composed  of  an  iron 
rod  GH  passing  through  the  knife-edge  to  which  it  is  held 
by  the  thumb-screw  N.  This  allows  for  the  adjustment 
cf  the  length  and  therefore  of  the  period  of  the  pendulum 
within  certain  limits.  Through  a  slot  near  the  upper  end 
of  the  pendulum  projects  a  slender  arm  RS  carrying 
the  electromagnet  T,  capable  of  holding  a  small  steel 
ball.  Near  the  bottom  is  a  metal  hook  K  which  engages 
with  a  metal  strip  on  the  end  of  a  lever  L  so  fastened  as 
to  hold  the  pendulum  about  10  or  15  degrees  from  the 


140 


ELEMENTARY  PRACTICAL  MECHANICS 


vertical.  By  raising  this  lever  the  pendulum  is  started 
and  the  electric  circuit  through  -the  magnet  is  broken  at 
the  same  instant,  therefore  the  pendulum  and  the  ball 
begin  their  motions  simultaneously. 

The  magnet  is  set  in  such  a  position  that  the  ball  would 
fall  exactly  along  the  face  of  the  pendulum,  if  the  latter 
were  at  rest  in  its  vertical  position,  and  hence,  if  ball  and 
pendulum  start  together,  they  will  strike  when  the  pendu- 
lum has  made  exactly  half  a 
swing.  It  is  evident  theretore, 
that  the  time  of  fall  can  be  found 
by  determining  the  period  of  the 
pendulum.  Where  the  ball  strikes 
it  leaves  a  mark  about  an  eighth 
of  an  inch  long  on  a  chalked  piece 
of  black  vulcanite  carried  on  the 
face  of  the  pendulum.  The  dis- 
tance from  the  electromagnet  to 
this  mark  is  the  fall  correspond- 
ing to  a  half  vibration. 

The  arrangement  of  the  appa- 
ratus is  such  as  to  allow  all  the 
necessary  adjustments  to  be  made 
quite  readily  and  exactly.  By 
using  a  good  stop-watch  and  tim- 
ing a  long  series  of  vibrations, 
the  period  of  the  pendulum  can 
be  found  very  closely.  The  chief 

source  of  difficulty  is  in  determining  the  exact  point  where 
the  ball  strikes  the  chalked  surface.  As  may  be  seen  from 
the  data  given  below,  these  points  vary  in  position  by  not 
more  than  2  cms.  in  about  250,  which  is  considerably  less 
than  one  per  cent,  while  in  most  cases  the  difference  is  only 
a  few  millimeters.  This  close  agreement  would  seem  to 


tod  Rubber  > 
Chalked 


FIG.  126. 


ELEMENTARY   PRACTICAL  MECHANICS 


141 


indicate  also  that  the  error  to  be  anticipated  from  the  fact 
that  the  ball  in  falling  will  not  necessarily  pass  over  an 
exactly  vertical  path,  thus  meeting  the  pendulum  at  times 
a  little  short  of  its  middle  position,  at  others  a  little  beyond, 
is  small.  Thus  by  averaging  a  number  of  readings,  a  prob- 
able value  may  be  found  which  is  not  only  quite  close  enough 
for  the  purposes  for  which  the  apparatus  is  designed  but 
which  approaches  very  closely  the  true  value. 

The  following  table  shows  the  results  which  the  appa- 
ratus is  capable  of  giving: 


Distance 
of  Fall, 

s. 

Average, 
& 

Period  of 
Pendulum. 

t. 

t*. 

248.2 

247.3 

248.2 

249.0 

248.0 
248.5 

247.7 

1.424 

.712 

.507 

247.4 

246.5 

247.0 

247.0 

Therefore,  by  the  general  forumla,  S  —  ^at2^  we  obtain 
0  =  977  cms/sec2,  a  result  within  about  one-third  per  cent 
of  the  accepted  value  980  cms /sec2.  The  value  977  here 
obtained  is,  of  course,  that  of  a  body  falling  in  air  instead 
of  in  a  vacuum. 


142  ELEMENTARY  PRACTICAL  MECHANICS 

PROBLEMS 

1.  If  a  pendulum   of  length   30  in.   has  a  period   .885 
second,  what  is  the  period  of  one  20  in.  long? 

2.  If  the  period  of  a  certain  pendulum  is  1.2  seconds, 
what  is  the  period  of  one  three  times  as  long? 

3.  A   pendulum   whose   length  is  6  makes  40  vibrations 
per  minute.     How  many  will  one  whose  length  is  2  make 
in  the  same  time? 

4.  A  certain  pendulum  is  found  to  make  one  vibration 
in  .8  second  at  a  place  where  the  value  of  g  is  31.8.     What 
will  be  its  period  at  a  place  where  g  is  32? 

5.  Calculate  the  length  of  a  pendulum  whose  period  is 
2  seconds. 

6.  The  period  of  a  pendulum  95  cms.  long  is  1.06  seconds. 
From  these  figures  find  the  value  of  g. 

7.  The  rope  carrying  the  car  of  a  mine  shaft  is  900  ft. 
long.     What  will  be  the  period  of  swing  as  a  pendulum? 

8.  A  rigid  pendulum  is  inclined  so  as  to  make  an  angle 
of  40°  with  the  vertical.     If  its  period  when  hanging  ver- 
tical was  .51  second,  what  will  it  be  when  thus  inclined? 

9.  A  rigid  pendulum  makes  2  vibrations  per  second  when 
hanging  vertical.     How  many  will  it  make  when  inclined 
at  an  angle  of  50°  to  the  vertical? 

10.  Compute  the  length  of  a  seconds  pendulum  (one  hav- 
ing a  period  of  one  second),  at  a  place  where  g  is  980.3  cms. 

11.  A  clock  having  a  pendulum  40  cms.  long  keeps  correct 
time.     The  pendulum  is   lengthened   to  40.2    cms.     How 
many  seconds  will  the  pendulum  lose  in  one  day? 


CHAPTER  IX 
FORCES  PRODUCING   MOTION 

IT  has  been  stated  that  a  force  may  produce  either  or 
both  of  two  effects  upon  a  given  body:  (a)  Change  in 
its  state  of  motion;  (b)  change  of  form.  In  our  study  of 
forces  thus  far,  we  have  assumed  a  "  rigid  body,'1  i.e.,  one 
in  which  any  change  of  form  which  may  occur  is  too 
slight  to  materially  affect  the  particular  conditions  with 
which  we  were  concerned,  and  have  confined  our  atten- 
tion solely  to  those  cases  in  which  the  forces  were  so 
balanced  that  the  body  considered  was  maintained  un- 
changed in  its  existing  state  of  motion,  i.e.,  continued  at 
rest  or  in  uniform  motion.  For  this,  we  found  it  necessary 
that  for  every  force  which  would  produce  displacement  in 
any  given  direction  there  should  be  an  exactly  equal  force 
in  the  opposite  direction,  and  for  every  torque  which 
would  cause  the  body  to  rotate  in  one  direction  about  a 
given  axis,  there  should  be  an  equal  torque  about  the 
same  axis,  but  in  the  opposite  direction. 

In  this  chapter  we  again  assume  a  rigid  body,  but  have 
to  deal  with  those  cases  in  which  the  force  in  any  direc- 
tion is  not  exactly  neutralized,  and  therefore  change  of 
motion  occurs  in  the  direction  of  the  unbalanced  force; 
or  in  which  the  torque  is  not  balanced  and  therefore  the 
rotating  body  increases  or  decreases  its  speed.  In  other 
words,  we  have  to  consider  all  those  cases  in  which  a  body 
is  acquiring  velocity,  is  coming  to  rest,  or  is  changing  the 

143 


144  ELEMENTARY  PRACTICAL  MECHANICS 

direction  of  its  motion.  We  are  to  investigate  the  laws 
which  govern  such  changes,  and  are  to  determine  the 
equation  which  shall  enable  us  to  compute  the  force  re- 
quired to  increase  or  decrease  the  speed  of  a  given  body 
by  a  stated  amount  in  a  stated  time,  or  the  time  required 
for  a  force  to  produce  or  destroy  a  given  velocity. 

79.  Meaning  of  the  Term  Mass.— It  is  a  matter  of  com- 
mon experience  that  it  is  more  difficult  (i.e.,  requires  more 
force)  to  set  in  motion  a  loaded  car  than  an  empty  one, 
to  stop  a  heavy  ball  rolling  along  than  to  stop  a  lighter 
one  moving  with  the  same  velocity,  etc.  To  take  an 
ideal  case  not  capable  of  perfect  experimental  proof,  con- 
sider two  blocks  of  the  same  shape  and  size,  one  of  wood 
and  one  of  lead,  resting  on  a  perfectly  smooth  horizontal 
table.  Attach  two  coiled  springs  exactly  alike,  one  to 
each  block,  and  pull  horizontally  on  each  so  that  the  two 
springs  will  be  extended  exactly  equal  amounts.  Each 
block  will  begin  to  move  in  the  direction  of  the  applied 
force,  but  the  block  of  wood  will  move  faster  than  the 
block  of  lead,  i.e.,  will  have  a  greater  acceleration.  Here 
equal  forces  act  on  two  bodies  under  identical  conditions. 
Why  are  the  resulting  accelerations  unequal?  Because 
the  lead  has  more  matter  in  it,  or  to  use  the  exact  term, 
the  lead  has  a  .greater  Mass. 

In  general,  it  is  a  result  of  our  experience  in  attempt- 
ing to  set  bodies  in  motion,  that  where  there  is  more 
matter  to  be  moved  there  will  always  be  less  motion,  i.e., 
less  velocity  acquired  in  a  given  time  under  the  action  of 
the  same  amount  of  force.  This  may  be  put  more  formally 
by  saying  that  under  the  influence  of  a  given  force,  a  body's 
acceleration  is  less  in  proportion  as  the  amount  of  matter 
in  the  body  is  greater. 

It  is  this  fact  which  we  recognize  by  saying  that  matter 
has  inertia. 


ELEMENTARY  PRACTICAL  MECHANICS 


145 


Definition. — The  amount  of  matter  in  a  body  is  called 
its  Mass.  Thus,  in  the  illustration  of  the  blocks  of  wood 
and  lead  just  given,  we  say  that  the  lead  block  has  a 
smaller  acceleration  because  it  has  a  greater  mass. 

The  inertia  of  a  body  exhibited  when  an  attempt  is 
made  to  change  its  velocity,  has  already  been  described 
as  one  of  the  forms  of  reaction  against  which  applied 
forces  act. 

80.  Relation  between  Force  and  Mass. — The  general 
relation  of  the  force  required  to  produce  motion  to  the 
quantity  of  matter  moved,  and  the 
rate  at  which  velocity  is  to  be 
acquired,  was  suggested  in  the  pre- 
ceding article. 

The   exact   law  of  this  relation 
ship  may  be  illustrated  by  the  fol- 
lowing simple  experiment: 

W  is  a  light  wheel  over  which 
weights  A  and  B  are  hung  by  a 
light  flexible  cord.  (Fig.  127.)  We 
will  assume  that  there  is  no  friction 
and  that  the  masses  of  wheel  and 
cord  are  so  small  that  they  may 
be  neglected.  Let  A  and  B  be 
brass  blocks  of  the  same  size  and 
shape,  each  45  grams,  and  let  R  be  a  third  block  of  10 
grams.  When  released  B  and  R  will  descend,  and  A  will 
ascend  with  accelerating  motion,  and  by  observing  the  dis- 
tance the  system  moves  in  a  given  number  of  seconds  we 
may  calculate  the  acceleration. 

Suppose  the  system  moves  196  cms.  in  2  seconds.     The 

196 
average  velocity  is  — =98;  the  final  velocity  98X2  =  196, 

and  the  acceleration  is  98. 


FIG.  127. 


146  ELEMENTARY  PRACTICAL  MECHANICS 

Now  add  masses  of  45  grams  to  each  side.  The  accelera- 
tion will  be  less,  and  a  second  extra  mass  R'  must  be  put 
on  in  order  that  acceleration  may  again  be  98. 

We  may  tabulate  our  facts  as  follows: 

Force.  Total  Mass.  Accel. 

Case  1  10  100  98 

2  10  190    less  than  98 

3  20  200  98 

From  our  data  we  see  that  in  order  to  produce  the  same 
acceleration  the  force  required  must  be  greater  in  proportion 
as  the  total  mass  moved  is  greater. 

To  investigate  our  problem  still  further  we  may  proceed 
thus:  Let  A  and  B  be  equal,  and  each  45  grams,  and  let 
R  consist  of  three  equal  pieces  each  5  grams.  Put  all  three 
on  the  side  with  B  and  find  by  test  as  before  the  accelera- 
tion; suppose  it  is  found  to  be  140.  Now  remove  one  of 
the  pieces  from  B  and  put  it  on  A.  The  mass  moved  is 
the  same  as  before,  but  the  moving  force  is  now  5  grams 
instead  of  15  grams.  We  shall  find  that  the  resulting 
acceleration  will  be  less — will  be,  in  fact,  46.6  or  J  of  140. 

Force.  Total  Mass.  Accel. 

Case  1  15  105  140 

2  5  105  46.6 

We  have  therefore  found  that  the  acceleration  given  to 
a  certain  body  depends  upon  the  forces  applied,  and  that 
if  the  force  acting  is  increased  the  acceleration  will  be  increased 
in  the  same  proportion. 

To  summarize  the  results  of  our  experiment  in  a  more 
condensed  and  mathematical  form,  we  may  use  the  follow- 
ing symbols:  /=  force,  ?^  =  mass,  a  =  acceleration. 


ELEMENTARY  PRACTICAL  MECHANICS  147 

From  the  first  part  of  our  experiment,  for  a  constant 
acceleration,  force  varies  as  mass,  or 


From  the  second  part,  for  the  same  mass,  the  acceleration 
produced  varies  as  the  force  applied,  or 


Therefore, 

This  is  a  perfectly  general  expression  of  the  natural 
law  which  applies  whenever  a  body  is  set  in  motion  or 
when  the  motion  of  a  moving  body  is  modified  in  any 
way.  . 

81.  Relation  of  Certain  Units  of  Mass,  Force,  and  Accel- 
eration. So  far  we  have  shown  merely  that  forces  are 
proportional  to  the  masses  moved,  and  the  accelerations 
produced,  or,  in  general,  fee  ma.  But  this  expression  in 
its  present  form  does  not  enable  us  to  determine  the  amount 
of  force,  measured  in  recognizable  force  units,  that  will  be 
required  to  impart  a  desired  acceleration  to  a  particular 
mass.  Before  such  computation  is  possible,  the  expression 
must  be  changed  to  the  equational  form,  f=ma,  or  "  force 
equals  mass  X  acceleration  ;"  and  before  such  an  equation 
can  be  written,  we  must  choose  units  of  force  and  mass 
which  will  make  the  equation  true  numerically. 

In  our  common  English  units  the  unit  mass  is  called  the 
pound.  The  force  of  gravity  drawing  this  mass  toward 
the  earth  is  also  named  a  pound.  It  is  thus  evident  that 
we  use  the  term  pound  in  two  quite  distinct  senses,  the 
first  indicating  a  mass,  that  is,  a  quantity  of  matter  exhibit- 
ing a  certain  inertia,  the  second  indicating  a  force.  Now, 


148  ELEMENTARY  PRACTICAL  MECHANICS 

if  we  let  a  pound  mass  fall  it  is  acted  on  by  a  force  of  a 
pound,  and  we  know  that  the  resulting  acceleration  is 
0  =  32  ft/ sec  in  one  second.  Thus  it  is  not  true  that 

Force  (in  pounds)  =  (mass  pounds)  X  acceleration  in  feet  and 
seconds,  would  for  this  give  1  =  1X32.  In  other  words, 
we  may  not  use  the  expression  pound  in  two  different 
senses  in  the  same  equation. 

But  suppose  we  elect  to  use  the  unit,  pound,  as  a  weight 
or  force  unit,  and  then  adopt  such  a  unit  of  mass  that  our 
equation  will  be  numerically  true.  Such  a  mass  unit  will 
evidently  be  the  mass  of  a  body  weighing  approximately 
32  pounds  because  the  weight  of  a  body  gives  to  its  mass  an 
acceleration  of  approximately  32" f if  sec2.  In  other  words, 
we  will  use  a  mass  unit  wrhich  will  satisfy  our  equation  if 
we  divide  the  weight  of  a  body  in  pounds  by  32;  and  our 
equation  may  now  be  written, 

1  Ib.  (force)  ==• — '—- —   -  (mass)X  acceleration  in  ft/sec2, 
oZ 

which  is  true. 

If,  therefore,  we  express  the  force  F  acting  upon  a 
body  in  gravity  units  or  pounds,  the  acceleration  a  with 
which  the  body  is  caused  to  move  in  feet  per  second  in 
one  second,  and  the  mass  of  the  body  in  terms  of  its  weight 
W  in  pounds  divided  by  g  (i.e.,  32  ft/sec2),  we  may 
express  the  true  relation  between  the  three  quantities  by 
the  general  equation, 


F=— a. 
9 


ELEMENTARY  PRACTICAL  MECHANICS  149 

This  is  the  equation  commonly  used  by  the  engineer 
who  finds  it  more  convenient  to  use  the  pound  as  a  unit 
of  force,  and  who  therefore  derives  a  unit  of  mass  to  satisfy 

.  weight 

the  equation  /  =  ma,  by  expressing  mass  in  terms  of  -  . 

9 

If  we  use  F  to  mean  the  force  in  grams,  W  the  weight  in 
grams,  and  g  the  acceleration  in  centimeters  per  second  in 
one  second  (980),  we  have  the  same  equation  again,  a 
must,  of  course,  always  be  expressed  in  the  same  units 
as  g. 

Illustrations.  —  (a)  What  acceleration  will  result  if  a  pull 
of  80  pounds  acts  on  a  body  weighing  640  pounds?  Here 
F  =  80,  F=640,  and 

640 
80  =  -Xa,       .    ._,  ..,      ,     .. 

whence  a  =  4  ft/  sec2. 

(6)  A  body  weighing  2940  grams  is  acted  on  by  a  con- 
stant force  and  an  acceleration  of  8  results.  Find  the 
force. 

2940 


jT  =  24  grams. 

These  two  units  of  force,  the  pound  force  and  the  gram 
force,  are  sometimes  called  Gravitational  Units  of  Force. 

82.  The  Dyne.  —  Assume  a  unit  of  force  just  sufficient 
to  give  a  mass  of  one  gram  an  acceleration  of  1  cm/  sec2. 
Evidently  this  is  less  force  than  the  force  of  gravity  on 
the  gram  mass,  for  the  latter  would  give  it  an  accelera- 
tion of  980  cms/sec2.  But  adopt  this  force  as  a  new  unit 
of  force,  then  in  this  case, 


150  ELEMENTARY  PRACTICAL  MECHANICS 

/  (force)  =  1,  m  (mass)  =  l,  and  a  (acceleration)  =  1,  and  we 
may  write, 


an  equation  numerically  correct  for  these  units  by  defini- 
tion.    This  new  unit  of  force  is  called  a  Dyne. 

Definition.  —  A  dyne  is  the  force  which,  acting  on  a  mass 
of  1  gram,  will  give  it  a  velocity  of  1  cm.  per  second  in  1 
second. 

The  dyne  is  seen  to  be  —  —  of  a  force  of  1  gram,  or  more 

strictly, 

1  gram=0  dynes. 

Illustrations.  —  (a)  What  force  acting  on  a  mass  of  800 
grams  will  in  5  seconds  give  it  a  velocity  of  600  cms.  per 
second? 

The  acceleration  =-—  =  120  cms/sec2. 
5 

/=  800  XI  20. 
=96,000  dynes. 

(6)  If  a  force  of  10,000  dynes  acts  on  a  mass  of  200  grams 
for  3  seconds,  what  velocity  will  it  produce? 

10,000  =  200  Xa, 

a  =50  cms  /sec2, 
50X3  =  150  cms.  per  second. 

(c)  With  what  force  in  dynes  is  a  mass  of  60  grams 
drawn  toward  the  earth  at  a  place  where  g  =  980.5? 


ELEMENTARY  PRACTICAL  MECHANICS  151 

The  force  is  60  grams  (weight), 

1  gram  =  980.5  dynes, 
60X980.5  =  58,830  dynes. 

(d)  A  force  of  90,000  dynes  is  found  to  produce  an 
acceleration  =  1800  on  a  certain  mass.  Find  the  mass. 

90,000  =  1800m, 
ra=50  grams. 

In  a  similar  manner  we  might  choose  a  force  which, 
acting  on  a  mass  of  1  pound,  would  give  it  an  acceleration 
of  1  ft /sec2.  This  force  would  be  only  ^  of  a  force  of 
one  pound. 

Calling  this  new  unit  of  force  by  a  new  name,  Poundal* 
we  again  have  the  equation, 

f=ma, 

where  /=poundals,  m= pounds  mass,  and  a  —  acceleration 
in  feet  and  seconds. 

The  dyne  and  poundal  are  called  Absolute  Units,  because 
their  values  do  not  depend  on  the  value  of  g,  which  varies 
from  place  to  place  on  the  earth's  surface  and  at  points 
below  or  above  the  surface. 

83.  Mass  and  Weight. — The  foregoing  discussion  sug- 
gests the  distinction  between  the  mass  of  a  body  and  its 
weight. 

*  NOTE. — We  shall  seldom  have  occasion  to  use  poundals,  since 
in  all  engineering  work  with  the  English  units  forces  are  more 
conveniently  expressed  in  pounds.  The  dyne,  however,  is  fre- 
quently used  in  practical  mechanics,  especially  in  electrical  and 
chemical  calculations. 


152  ELEMENTARY  PRACTICAL  MECHANICS 

Mass  means  quantity  of  matter  measured  by  its  inertia. 

Weight  means  strictly  the  force  due  to  gravity  with 
which  the  body  is  attracted  to  the  earth. 

Since  the  quantity  of  matter  does  not  change  when  a 
body  is  moved  from  one  place  to  another,  the  mass  is 
constant.  But  the  weight  may  change,  indeed,  usually 
does  change.  It  is  less  at  the  equator  than  near  the  poles, 
for  instance.  Also  on  the  moon  the  weight  of  what  we 
call  a  pound  of  iron  would  be  much  smaller  than  on  the 
earth,  but  its  mass  would  be  the  same,  for  the  same  force 
applied  to  it  would  produce  the  same  acceleration. 

At  the  center  of  the  earth  a  body  would  have  the  same 
mass  as  elsewhere  but  no  weight  at  all,  but  if  a  mass 
were  in  motion  there,  the  same  force  would  be  required 
to  stop  it  as  at  the  earth's  surface. 

84.  Summary. — The  essential  meaning  of  the  term 
Force  is  that  which  gives  or  tends  to  give  accelerated 
motion  to  a  mass,  and  it  is  measured  by  the  product  of 
mass  and  acceleration.  If  a  body  has  no  acceleration  (is 
in  equilibrium)  it  does  not  necessarily  follow  that  there  is 
no  force;  it  may  mean  that  there  are  two  equal  and  oppo- 
site forces  acting,  or  any  system  of  forces  such  that  they 
neutralize  each  other.  Any  one  of  the  forces  alone  would 
produce  motion,  the  reaction  being  in  the  inertia  of  the 
body.  Two  systems  of  force  measurement  are  in  common 
use.  In  one  the  -terms  pound,  gram,  etc.,  are  taken  as 
names  of  force  units.  In  this  system  mass  is  expressed 
by  the  derived  unit,  weight,  This  system  is  called  the 
gravitational  system.  In  the  other  system  the  pound, 
gram,  etc.,  are  units  of  mass.  A  unit  of  force  is  then  derived 
by  assuming  as  unit  force  that  force  which  will  impart 
unit  acceleration  to  unit  mass. 

The  table  below  shows  the  relations  between  units. 


ELEMENTARY  PRACTICAL  MECHANICS  153 


Force  Units. 

Mass  Units. 

Accel.  Unit,  a. 

Equations. 

Gravitation. 

Absolute. 

F=  pounds 
/=poundals 
F=  grams 
/=  dynes 

TF\lbs.)     W 

1  ft/sec.2 
t( 

1  cm/sec.2 
tt 

'-5- 

f=ma 
f=ma 

9            32 
m  =  lbs. 
W(gm.)      W 

P        Wn 

g          980 
77i  =gm. 

F    980° 

85.  Motion  of  a  Body  on  a  Smooth  Incline. — In  Fig.  128 
OA  represents  the  total  pull  of  gravity  W  on  a  body  0. 
Then  OB  and  OC  represent  the 
components  of  W  parallel  to 
and  perpendicular  to  the  face 
of  the  incline,  if  d  is  the  angle 
between  the  plane  and  the  hori- 
zontal then  OB  =  W  sin  d,  or 


V 

SB"  A 

FIG.  128. 
If  the   surface   SR    is    smooth 

(i.e.,  if  there  is  no  friction),  the  component  OC  does  not 
affect  the  motion,  and  the  moving  force  is  OB  tending 
to  give  motion  down  the  plane.  Since  the  full  force  of 


-^ 

O/i 


gravity  W  will  give  an  acceleration  0,  the  force  OB 

r>/77 

will  give  an  acceleration  a=gX-^=g  sin  d. 

Olt 

86.  Acceleration  Diagrams  and  the  Solution  of  Problems. 

—  When  a  body  is  made  to  move  with  accelerated  motion 
in  any  direction,  the  force  F\  in  the  direction  of  motion 
must  always  be  greater  than  the  force,  F2,  opposing  motion. 


154  ELEMENTARY  PRACTICAL  MECHANICS 

The   difference,    FI  —  F%,    represents   the   unbalanced  force 
causing   acceleration.     Opposed    to    this    unbalanced    force 
is  an  equal  reaction  due  to  inertia,  or  FI  — F2  =  inertia  re- 
weight  of  body  or  bodies  moved 
action  = X  acceleration.    If 

g 

we  include  this  inertia  reaction  as  an  equivalent  force  opposing 
change  of  motion,  we  may  write  the  general  equation  for 
accelerated  motion: 

Sum  (all  actions  in  the  directions  of  the  change  of  velocity) 
—  Sum  (all  reactions  in  the  opposite  direction). 

Thus  suppose  a  body  A  is  made  to  slide  with  increasing 
speed  over  a  rough  horizontal  surface,  Fig.  129,  bv  a  pull 
FI. 

Acceleration  occurs  in  the  direction  of  FI;  reactions 
are  due  to  friction  and  inertia.  Therefore 

/.       wt.  A 

FI  =  friction  +  inertia  reaction!  i.e.. Xacc 

\  9 

If  body  A,  already  in  motion,  is  being  brought  to  rest 
by  friction,  our  "  acceleration  diagram"*  will  be  as  in 
Fig.  130.  Friction  is  decreasing  the  velocity  (i.e.,  pro- 

*NOTE:  The  student  must  Dot  confuse  what  is  here  called  the 
"acceleration  diagram'7  with  the  force  diagram  of  a  free  body 
in  equilibrium.  The  moving  body  is  not  in  equilibrium.  In 
reality,  the  force  F,  which  causes  A  to  move  (Fig.  129),  comes 
from  some  second  body  B.  The  inertia  reaction  of  A  is  there- 
fore really  exerted  on  B  and  not  on  body  A,  as  shown.  Fig.  129  is 
therefore  not  a  force  diagram  for  body  A.  The  student  will,  how- 
ever, find  such  " acceleration  diagrams"  helpful  in  more  com- 
plicated cases  of  accelerated  motion,  because  they  help  to  create 
a  clear  idea  of  the  true  conditions,  and  tend  to  prevent  the  omis- 
sion of  any  forces  or  reactions  affecting  the  motion  of  the  given 
body. 


ELEMENTARY  PRACTICAL  MECHANICS 


155 


ducing  a  negative  acceleration),  and  the  inertia  of  A  is 
tending  to  continue  its  motion.     Therefore, 

wt.  A 
Friction = inertia  reaction  = X  ace. 


Acceleratfon  v 
'inertia 
(action 


_^  Acceleration 


Reae 


JEriction 


FIG.  129. 

In  general,  in  the  solution  of  problems,  proceed  as  follows: 

1.  Determine  the  direction  of  the  change  in  velocity. 

2.  Construct  an    "  acceleration    diagram/'    showing    all 
actions  in  the  direction  of  the  change  in  velocity,  and  all 
reactions    in    the    opposite    directions.     Always    include 
among  the  latter,  one  due  to  inertia. 

3.  Substitute  the  values  for  these 
actions  and  reactions  in  the  general 
equation  and  solve. 

ILLUSTRATIONS.  Example  1. — 
Weights  of  90  and  80  pounds  hang 
by  a  cord  over  a  pulley  (Fig.  131). 
Find  the  acceleration  of  the  system, 
if  friction  and  the  mass  of  cord  and 
wheel  may  be  neglected. 

Here  the  total  weight  moved  W 
is  80  +  90  =  170  pounds.  The  force  in 
the  direction  of  the  acceleration  is 
90  pounds;  opposing  reaction  on  the 
moving  system  are  the  gravity  pull  of  80  pounds  and  the 
inertia  reaction.  Therefore, 


from  which, 


FIG.  131. 


32' 

a  =  1.88  ft/sec2. 


156 


ELEMENTARY  PRACTICAL  MECHANICS 


Inertia 


Example^.— A  (Fig.  132)  =500  pounds,  and  P  =  30  pounds. 

If  there  is  no  friction,  find 
the  acceleration  and  the 
time  required  for  the  sys- 
tem to  move  6  ft. 

Here  total  weight  W  to 
be  moved,  =500  +  30  =  530 
pounds.  The  gravity  pull 
on  A  is  borne  by  the  sur- 
face and  the  gravity  pull 
•pIG  ^32  on  P  alone  causes  motion. 

Therefore, 


From  laws  of  accelerated  motion,  when  S  =  6  and  a  =  1.81, 

12 


2.58  seconds. 


Example  3. — A  car  weighing  1800  pounds 
is  drawn  vertically  up  a  mine  shaft.  In 
starting  the  car  the  acceleration  is  found 
to  be  20  ft /sec2.  Find  the  tension  in  the 
cable  during  start. 

The  "  acceleration  diagram  "  for  the  car  is  here  as  shown 
in  Fig.  133. 


FIG.  133. 


ELEMENTARY  PRACTICAL  MECHANICS  157 

The  force  causing  motion  is  the  pull   T  in  the  cable; 
the  opposing  actions  are  gravity  pull,  W,  and  inertia, 

F  =  1800  pounds; 

1800 

Inertia  reaction  =  —  —  X  20. 
32 

(a)  Therefore,       T  =  1800+—  —  X20. 

32 

From  which,  T  =  2925  pounds. 

(b)  Another  solution  may  be  given,  thus: 

TF  =  1800  pounds;   and  a  =  20.     Let  r=pull  in  cable  to 
give  acceleration  20;  then 


Add  to  this  1800  pounds,  the  car's  weight,  and  so  obtain 
the  tension,  2925  pounds  necessary  to  hold  car  up  and 
also  give  the  required  upward  acceleration. 

(c)  Still  another  solution  is:  Let  IF  =  1800  and  a  =  20  + 
32=*52,  that  is,  an  acceleration  upwards  equal  to  the 
downward  acceleration  of  gravity  +  20.  Then 


T=2925. 


158 


ELEMENTARY  PRACTICAL  MECHANICS 


Example  4—W  (Fig.  134)  =90  Ibs.,  P=40  Ibs.,  friction 
between  W  and  the  plane  =  3  Ibs., 
of  pulley  =  2  Ibs.  Find  the  direc- 
tion and  the  amount  of  the  accel- 
eration and  the  tension  of  cord 
both  sides  of  pulley. 

The  force  tending  to  move  the 
system  so  that  W  will  move  down 
the  plane  is  90  sin  35°  =  51.  7  Ibs. 
The  forces  opposing  this  motion 

are  P=40  and  friction  =  5  Ibs.  The  force  down  the  plane 
is  greater  and  therefore  change  of  motion  will  be  down 
the  plane.  The  total  weight  to  be  accelerated  =90  +40  = 

130 

130,  and  the  inertia  reaction  •—  r-Xa. 

32 

Therefore,  we  may  write  the  equation, 


FIG.  134. 


51.7=40  +  5+_ 


ISO 
32 


From  which,  a  =1.65  (downwards). 

To  find  tension  T\  in  cord  between  W  and  pulley, 
consider  W  alone.  The  acceleration  diagram  is  as  in 
Fig.  135(a). 


(a) 


(6) 
Fia.  135. 


ELEMENTARY  PRACTICAL  MECHANICS  159 


Therefore,          51.7= 


x  1.65. 


or,  1 1  =  44  Ibs. 

Then,  from  pulley  (Fig.  135(6) )  we  have,  neglecting  mass 
of  the  pulley, 


T2  =  42.1  Ibs. 

Or,  T2  may  be  found  from  the  equation  for  P  alone  (Fig. 

135(c)  ), 

'     40 


Example  5.  —  A  weight  £  =  1000  Ibs.  is  supported  by  a 


Inertia 
reaction  J 


IDOO* 


FIG.  136. 


FIG.  137. 


combination  of  pulleys  as  in  Fig.  136.     Load  A  =550  Ibs. 
With  what  acceleration  will  the  combination  move  and 


160  ELEMENTARY  PRACTICAL  MECHANICS 

what  will  be  tension  in  the  rope?  (Neglect  friction  and 
weight  of  pulleys.) 

It  will  be  seen  from  the  figure  that  motion  of  A  is  twice 
that  of  B.  Therefore  acceleration  of  A  will  be  twice  the 
acceleration  B. 

Let  a  =  acceleration  of  B.  Then  2a  =  acceleration  of  A. 
Let  T  =  tension  of  rope;  since  there  is  no  friction  the 
tension  everywhere  in  the  rope  is  the  same.  The  diagrams 
for  A  and  B  are  as  in  Fig.  137. 

Therefore  we  may  write  the  equations: 


550 
(1) 


(2)          2T 

o 

These  simultaneous  equations  may  be  solved  for  T  and 
a  in  the  usual  way. 


PROBLEMS 

1.  A  constant  force  acting  upon  a  mass  of  15  grams  for 
4  seconds  gives  it  a  velocity  of  20  cms.  per  second.     Find 
the  force. 

2.  A  man  pushes  steadily  against  a  car  on  a  level  track 
for  1  minute  and  thereby  produces  a  velocity  of  12  ft.  a 
second.     A  horse  in  the  same  time  can  produce  a  velocity 
of  20.5  ft.  a  second.     Compare  the  force  exerted  by  the 
man  with  that  exerted  by  the  horse. 

3.  A  force  A  can  produce  a  velocity  of  100  in  5  seconds, 
on  a  certain  mass.     A  force  B  can  produce  on  the  mass 
a  velocity  of  50  in  25  seconds.     How  many  times  as  large 
as  B  is  the  force  A. 


ELEMENTARY  PRACTICAL  MECHANICS  161 

4.  If  gravity  on  the  earth  gives  all  bodies  an  acceleration 
of  981  cms/sec2,  and  on  the  moon  an  acceleration  of  163  5 
cms /sec2,    compare   the  force   of  gravitation   on   the  two 
spheres. 

5.  A  mass  of  2  Ibs.  at  rest  is  struck  and  starts  off  with 
a  velocity  of  10  ft.  a  second.     Assuming  the  time  during 
which  the  blow  lasts  to  be  -^  of  a  second,  find  the  average 
force  acting  on  the  mass. 

6.  A  car  slides  down  a  smooth  incline  with  slope  1  ft. 
drop  in  5  ft.  length  of  track.     Find  (a)  acceleration;    (b) 
velocity  acquired  in  8  seconds;    (c)  distance  traversed  in 
5  seconds. 

7.  Sand  will  stop  a  certain  projectile  fired  into  it  in  y1^  of 
a  second.     Loam  will  stop  the  same  body  going  twice  as 
fast  in  |  of  a  second.     Compare  the  resistances  to  pene- 
tration offered  by  sand  and  loam. 

8.  What  velocity  would  a  train  acquire  in  half  a  minute 
on  a  3  per  cent  grade  if  there  were  no  friction  and  what 
distance  would  it  travel  in  that  time? 

9.  A  body  weighing  90  Ibs.  is  moving  at  the  rate  ot 
80  ft/sec.      What  force  expressed  in  pounds  will  stop  if 
in  2  seconds?     What  force  in  -^  of  a  second? 

10.  In  an  apparatus  such  as  shown  in  Fig.  131,  A  and 
B  weigh  10  Ibs.  and  9  Ibs.  respectively,  and  hang  by  a 
flexible  string  over  the  frictionless  pulley  W. 

(a)  Find  acceleration  of  the  system; 

(b)  Distance  traversed  in  two  seconds. 

11.  In  an  Atwood's  machine  the  weights  carried  by  the 
thread  are  6J  ounces  each.     The  friction  is  equivalent  to 
a  weight  of  ^  ounce.     When  the  "  rider/'  which  weighs 
1  ounce,  is  in  position,  what  will  be  its  gain  in  velocity 
per  second?     (See  Fig.  127.) 

12.  A  train  weighing   1000  tons  is  going  30  miles  an 
hour.     The    brakes    can    apply    a    force  =  10  tons    weight. 
If  on  a  level,  neglecting  friction  of  track,  etc.,  how  long 
will  it  be  before  the  train  will  stop?     How  far  will  it  go 
before  stopping? 


162  ELEMENTARY  PRACTICAL  MECHANICS 

13.  Answer  above  if  train  is  going  up  an  incline  of  1  in  20. 

14.  A  body  on  a  plane  rising  1  in   10  goes   100  ft.  in 
10  seconds.     Find  force  of  friction. 

15.  In  the  apparatus  shown  in  Fig.   132,  A  =  100  Ibs., 
P  =  5  Ibs.     No  friction.     How  fast  will  A  be  moving  after 
going  10  ft.  along  the  table? 

16.  If   A  =  900  Ibs.    and   friction   amounts   to   force    of 
8  Ibs.,  what  must*  be  the  weight  of  P  if  A  is  to  move  ^  ft. 
from  rest  in  i  second?     (See  Fig.  132.) 

17.  A  man  weighing  175  Ibs.  stands  on  the  floor  of  an 
elevator  which  is  descending  with  uniform  acceleration  of 
1  ft /sec2.     What  will  be  the  pressure  of  his  feet  on  the 
floor?     What  will  be  the  pressure  when  elevator  is  ascend- 
ing with  same  acceleration? 

18.  A  rope  winding  on  a  drum  hauls  a  car  vertically 
up  the  shaft  of  a  mine  with  an  acceleration  of  2  ft /sec2. 
If  car  weighs  1J  tons  find  tension  in  rope. 

19.  A  railroad  train  is  moving  on  a  level  at  the  rate  of 
30  miles  an  hour.    The  train  weighs  400  tons.     If  steam 
is  shut  off  what  force  due  to  the  brake  is  required  to  stop 
the  train  within  200  ft.     (Neglect  other  frictions.) 

20.  In  the  apparatus  of  Fig.  134,  CD  =  5  ft.,  C#=50  ft., 
W =9000  Ibs.,  P=40  Ibs.     Find  velocity  in  4  seconds. 

21.  A  bullet  weighing  2  ounces  going  at  rate  of  1000  ft/sec 
penetrates  15  in.  into  an  oak  post.    Find  average  resist- 
ance to  penetration  offered  by  post. 

22.  If,  in  diagram  Fig.   131,  B  is  18  gm.,  what  must 
be  the  mass  A  to  give  the  system  an  acceleration  of  40 
cms/ sec2? 

23.  If  in  Fig.  131  A  is  80  gm.  and  B  is  60,  find 

(a)  Acceleration; 

(b)  Tension  of  the  string. 

24.  If  in  Fig.  131  A  is  80  gm.  and  B  60  gm.,  and  A  is 
moving  upward  with  velocity  of  90  cms.  per  second,  how 
far  and  how  long  will  it  move  before  coming  to  rest? 


ELEMENTARY  PRACTICAL  MECHANICS  163 

25.  If  in  Fig.  131  A  =  6  and  B  =  5  Ibs.,  how  long  will  it 
be  before  the  velocity  will  be  300  ft/sec? 

26.  In  an  Atwood's  machine  the  weights  are  90  gms.  and 
80  gms.     The  system  moves  from  rest  for  4  seconds,  when 
the  string  breaks.     How  long  will  the  80-gm.  weight  con- 
tinue to  rise,  and  how  far  will  it  rise?     (See  Fig.  127.) 

27.  A  platform  sustaining  a  weight  of   120  Ibs.   is  de- 
scending with  a  uniform  acceleration  of  4  ft /sec2.     Find 
the  pressure  of  the  weight  on  the  platform. 

28.  Find  the  pressure  between  weight  and  platform  in 
the  case  above  if  the  platform  is  ascending. 

(a)  With  uniform  velocity  of  9  ft /sec. 

(b)  With  a  uniform  acceleration  of  10  ft /sec2. 

29.  In    an    inclined    plane,    slope    AC  =  20  ft.,    height 
CD  =  4  ft.     Body  weighing  100  Ibs.  rests  on  the  incline. 

(a)  Find  acceleration  along  AC. 
(6)  Find  pressure  on  AC. 

(c)  Find  time  to  pass  from  C  to  A. 

30.  In  the  inclined  plane  of  Fig.  134,  EC  =  13  ft.,  CZ>  = 
5  ft.,  TF  =  91  Ibs.,  P=100  Ibs.     Find 

(a)  Direction  of  motion  and  the  acceleration. 

(b)  Tension  in  the  string  during  motion. 

31.  In  the  incline  plane  of  Fig.  134,  TF  =  210  Ibs.,   EC= 
28  ft.,  CD  =  6  ft.;    force  of  friction  =  10  Ibs.     Find  mass 
ofP, 

(a)  To  produce  equilibrium. 

(b)  To  give  acceleration  of  8  ft/sec2  up  the  plane. 

32.  A   mass  of  80  gms.  slides   down   a   smooth   incline 
whose  height  is  ^  its  length  and  draws  another  mass  from 
rest,  over  a  distance  of  10  ft.  in  3  seconds  along  a  horizontal 
table  level  with  the  top  of  the  plane,  the  string  passing 
over  the  top  of  the  plane.     Find  the  mass  on  the  table. 

33.  A  body  is  projected  up  an  incline  rising  2  meters  for 
each  49  meters  of  track  with  an  initial  velocity  of  200 
meters  per  second.     Find  the  velocity  and  distance  at  the 
end  of  6  seconds. 


184  ELEMENTARY  PRACTICAL  MECHANICS 

34.  In  what  time  would  the  body  of  Prob.  33  come  to 
the  highest  point  it  will  reach  on  the  incline? 

35.  A  car  slides  down  an  incline  80  meters  long  and 
having   an   inclination    of   45   degrees   to    the   horizontal. 
Find   the   velocity   at   the   bottom   and   the  time   of  the 
descent. 

36.  An  elevator  loaded  with  passengers  weighs  1450  Ibs. 
It  is  counterbalanced  with  weights  weighing  950  Ibs.    With 
what  acceleration  would  the  elevator  descend  if  it  were 
free  to  fall  and  there  were  no  friction  on  the  pulleys  or 
guides? 

37.  A  weight  of  2000  Ibs.  is  supported  by  a  system  of 
pulleys,  as  indicated  in  Fig    136.     There  is  at  A  a  weight 
of  1200  Ibs.     Find  the  time  that  it  will  take  the  weight  A 
to  fall  100  ft. 

38:  A  N.  Y.,  N.  H.  &  H.  electric  locomotive,  weighing 
200  tons  and  hauling  a  200-ton  train,  exerts  an  average 
tractive  force  of  10,000  Ibs.  Neglecting  train  friction,  what 
time  will  be  required  to  get  up  a  full  speed  of  60  miles  an 
hour  from  rest? 

39.  What  time  would  be  required  by  the  train  of  Prob.  38 
if  on  a  1  per  cent  up  grade? 

40.  If  the  train  of  Prob.  38  starts  up  a  3  per  cent  grade 
at  a  speed  of  55  miles  per  hour,  what  will  be  its  speed  at 
the  end  of  a  half  mile? 

41.  If  the  locomotive  of  Prob.  38  is  required  to  get  up 
full  speed  in  200  seconds,  what  is  the  maximum  trailing 
load  it  can  haul? 

87.  Force  and  Acceleration  in  Rotary  Motion. — In  the 

preceding  paragraphs  of  this  chapter,  we  have  shown  that 
inertia  is  the  property  of  all  matter  by  reason  of  which 
it  offers  resistance  to  linear  acceleration;  that  inertia 
depends  solely  upon  mass;  and  that  the  fundamental 
relations  in  motion  of  translation  are  expressed  by  the 
equation,  force = mass  X  acceleration. 


ELEMENTARY  PRACTICAL  MECHANICS  165 

When  we  come  to  apply  tbsse  ideas  to  motion  of  rota- 
tion, we  find  that,  because  of  the  nature  of  such  motion, 
certain  other  factors  must  be  taken  into  account.  Thus, 
the  ability  of  a  force  to  produce  rotation  does  not  depend 
upon  its  amount  alone  but  also  upon  its  moment  arm. 
A  small  force  having  a  long  moment  arm  will  produce  as 
much  tendency  to  rotation  (i.e.,  torque)  as  a  greater  force 
having  a  correspondingly  short  arm.  The  resistance 
offered  by  a  body  to  acceleration  in  rotation  (i.e.,  its  rota- 
tional inertia),  does  not  depend  upon  mass  alone  but  also 
upon  the  way  in  which  the  mass  is  distributed  with  respect 
to  the  axis  of  rotation.  The  inertia  reactions  of  masses 
farther  out  from  the  axis  have  greater  moment  arms  and  there- 
fore react  with  greater  effect  in  opposing  rotation.  Thus  it 
is  a  matter  of  common  experience  that  a  wheel  with  its 
mass  mainly  in  its  rim  offers  much  greater  resistance  to 
starting  and  stopping  than  one  which  weighs  exactly  the 
same  but  which  has  its  material  concentrated  near  its 
axle.  Then,  also,  in  rotation,  the  linear  acceleration  given 
the  particles  of  the  body  must  be  greater  in  proportion  to 
their  distances  out  from  the  axis  of  rotation. 

If,  howrever,  we  express  the  tendency  to  produce  rota- 
tion in  terms  of  units  of  moment,  the  acceleration  as  angular 
acceleration,  which,  as  we  have  seen,  is  constant  for  all 
portions  of  the  body,  and  if  we  assume  that  a  body  offers 
unit  resistance  to  angular  acceleration  (i.e.,  has  unit  rota- 
tional inertia),  when  one  unit  of  moment  (unit  torque)  is 
required  to  give  the  body  unit  angular  acceleration,  we  may 
write  as  an  equation  expressing  the  fundamental  relations 
in  rotary  motion: 

MOMENT  OF  FORCE  (ALSO  CALLED  TORQUE)  =  ROTATIONAL 
INERTI AX  ANGULAR  ACCELERATION. 

In  this  equation  UNIT  MOMENT  OF  FORCE  =  a  force  of  1  lb., 
having  a  moment  arm  of  1  ft.,  a  force  of  1  gm.  having  an 


166  ELEMENTARY  PRACTICAL  MECHANICS 

arm  of  1  cm.,  a  force  of  1  dyne  having  an  arm  of  1  cm.,  etc., 
or  their  equivalents,  according  to  the  system  of  forces  and 
lengths  used.  These  may  be  called  simply:  One  pound- 
foot,  one  gram-centimeter,  one  dyne-centimeter,  etc. 

UNIT  ANGULAR  ACCELERATION  =  1  radian  per  sec2. 

Angular  acceleration  expressed  as  revolutions  per  sec2  X  2;r 
=  acceleration  expressed  as  radians  per  sec2,  since  there 
are  2x  radians  in  a  complete  circumference;  therefore,  the 
general  equation  for  rotary  motion  may  also  be  written: 

Torque  =  rotational  inertia  X  2rc  X  acceleration 
in  revolutions  per  sec2. 

88.  Moment  of  Inertia. — The  resistance  which  a  body 
offers  to  angular  acceleration   (i.e.,  its  rotational  inertia) 
is  commonly  called  its  moment  of  inertia. 

UNIT  MOMENT  OF  INERTIA*  is  the  moment  of  inertia  of  a 
body  to  which  unit  moment  of  force  will  impart  unit  angular 
acceleration. 

89.  Radius  of  Gyration. — Attention   has    already   been 
called  to  the  obvious   fact  that  in  rotary  motion  portions 

*  It  may  be  shown  that  the  moment  of  inertia  of  a  par- 
ticle is  equal  to  the  mass  of  the  particle  w,  multiplied  by  the 
square  of  its  distance  r  from  the  axis  of  rotation  or  mr2.  The 
moment  of  inertia  /  of  a  body  equals  the  sum  of  the  moments 
of  inertia  of  its  separate  particles,  or  /=sum  (mr2). 

To  compute  the  moment  of  inertia  of  a  body,  therefore,  it  is 
necessary  to  multiply  the  mass  of  each  of  its  particles  by  the 
square  of  its  distance  from  the  axis  and  then  to  find  the  sum  of 
all  such  products.  This  is  impossible  for  irregular  bodies  of 
varying  density,  and  in  such  cases  /  must  be  found  by  experiment. 
In  case  of  regular  homogeneous  bodies  the  moments  of  inertia 
with  respect  to  given  axes  may  be  calculated  by  the  methods 
of  the  calculus.  This  is  beyond  the  scope  of  this  text  and  the 
reader  who  is  interested  is  referred  to  more  advanced  treatises 
and  to  tables  of  moments  of  inertia,  etc. 


ELEMENTARY  PRACTICAL  MECHANICS  167 

of  the  rotating  body  at  different  distances  from  the  axis 
have  different  linear  velocities.  It  is  evident  upon  con- 
sideration, however,  that  there  must  be  a  point  in  every 
rotating  body  which  will  move  with  such  a  velocity  that  if 
the  entire  mass  of  the  body  could  be  concentrated  at  this 
point,  the  resistance,  which  the  body  would  offer  to  a 
torque  tending  to  change  its  angular  velocity,  would  be  un- 
changed. 

The  distance  from  the  axis  of  rotation  to  this  point  is 
known  as  the  radius  of  gyration  for  the  body. 

In  rotary  motion  we  may,  therefore,  when  more  con- 
venient, consider  the  entire  mass  of  the  body  as  having 
the  same  linear  velocity  as  a  point  situated  a  distance 
equal  to  the  radius  of  gyration  from  the  axis  of  rotation. 

The  value  of  the  radius  of  gyration  for  various  common 
forms  as  disks,  cylinders,  etc.,  is  given  in  most  practical 
handbooks.  In  the  case  of 
fly-wheels  having  a  heavy 
rim  it  is  customary  to  neg- 
lect the  mass  of  the  hub 
and  spokes  and  regard  the 
radius  of  gyration  as  the 
distance  from  the  axis  to 
middle  of  rim  as  r*o,  Fig. 
138. 

90.  Computation  of 
Torque  and  Rate  of  Increase 
or  Decrease  of  Speed  for 
Fly-wheels,  Shafts,  etc. — If  /  is  the  force  in  absolute  units 
which  causes  a  fly-wheel  to  rotate  and  d  the  moment  arm 
of  this  force  with  respect  to  the  axis  about  which  the 
wheel  turns  (see  Fig.  138),  then  fd= torque  which  produces 
rotation. 

If  M=mass  of  rim  and  ao  =  linear  acceleration  at  the 


168  ELEMENTARY  PRACTICAL  MECHANICS 

distance  r0  from  the  axis,  then  Ma0  is  a  measure  of  the 
inertia  reaction  of  the  rim,  and  Ma0r0  =  moment  of  this 
reaction  with  respect  to  the  axis.  Therefore 


This  is  the  fundamental  equation  for  the  relation  of 
torque,  mass,  and  acceleration  for  rotating  bodies  where 
radius  of  gyration  is  known. 

As,  however,  it  is  customary  to  express  velocity  of 
rotation  in  revolutions  per  second,  and  acceleration  of  the 
rotating  body  in  terms  of  revolutions  per  second  gained  or 
lost  each  second,  a  more  convenient  form  of  the  same 
equation  may  be  derived  as  follows: 

Let  ar  =  acceleration  in  revolutions  per  sec2.  Then 
acceleration  a0  in  radians  =  27rar,  and  linear  acceleration  = 
acceleration  in  radians  X  radius,  or 

ao  =  2;rroar, 

Substituting  this  value  for  a0  in  the  equation  above,  we 
have 


NOTE.  Since  moment  of  inertia  (or  rotational  inertia)  /= 
sum  (rar2),  and  the  whole  mass  M  of  a  body  may  be  regarded 
as  concentrated  at  a  distance  equal  to  the  radius  of  gyration 
r0  from  the  axis,  it  is  evident  that 


The  equations, 

Torque  =  moment  of  inertia  X  angular  acceleration,  or 
fd  =  1  X  acceleration  in  radian  /sec2,  and 


are  therefore  identical,  as  /=Jfr02,  and  2^Or  =  angular  acceleration. 


ELEMENTARY  PRACTICAL  MECHANICS 


169 


If  force  is  to  be  expressed  in  gravitational  units  F,  this 
expression  becomes, 


W 

— 

g 


This  is  the  equation  commonly  used  for  practical  com- 
putations for  fly-wheels/  shafts,  etc.,  where  the  radius  of 
gyration,  r0,  is  known.  Force  F  is  here  usually  the  differ- 
ence in  tension  between  tight  and  loose  sides  at  the  driving 
belt,  and  d  is  the  radius  of  the  belt  pulley. 

Example.  —  A  fly-wheel  whose  rim  weighs  8000  Ibs.  is 
started  from  rest  by  a  pull  of  1000  Ibs.  acting  at  the  cir- 
cumference of  a  belt 
pulley  of  12  in.  diam- 
eter. Outside  diameter 
of  fly-wheel  is  6  ft. 
6  in.,  thickness  of  rim 
6  in.  How  long  will 
it  take  to  reach  a  speed 
of  120  revolutions  per 
minute?  (Neglect  fric- 
tion.) 

Fly-wheels  are  usu- 
ally designed  to  have 
as  large  moment  of 
inertia  as  possible  for  given  conditions.  The  material  is 
therefore  mainly  in  the  rim.  Since  the  spokes  and  hub 
are  light  in  comparison  and  the  material  near  the  axis 
offers  relatively  little  resistance  to  angular  acceleration,  it 
is  customary  to  neglect  the  mass  of  spokes  and  hub  and  to 
take  as  the  radius  of  gyration  the  distance  from  center  of 
axis  to  middle  of  rim. 


1000  * 


FlG> 


170  ELEMENTARY  PRACTICAL  MECHANICS 

Radius  of  gyration  is  here  3  ft.     Mass  of  rim  -     -  units, 

{JA 

moment  arm  of  1000  Ibs.  force  £  ft.     Therefore,  from  our 

W 

expression  Fd= —  rQ22xar, 
y 

8000 

1000 Xi=-  -X3X3X2X3.14Xaccel.  in  revs,  per  sec2. 
o2 

Or, 

Accel.  =  .035  revolutions  per  sec2. 

120  revs,  per  min.  =2  revs,  per  sec.  to  be  gained. 

Therefore, 

2 

Time  required  =  — —  =  57  sec.  approx. 
.Ooo 


PROBLEMS 

1.  What  pull  in  a  belt  running  on  a  belt  pulley  20  in. 
diameter  will  be  required  to  increase  the  speed  of  a  fly- 
wheel from  120  r.p.m.  to  180  r.p.m.  in  10  seconds  if  weight 
of  fly-wheel  is  1600  Ibs.  and  its  radius  from  axis  to  center 
of  rim  is  2  ft.? 

2.  A  force  of  10  Ibs.  is  acting  on  a  fly-wheel  at  a  dis- 
tance of  3  in.  from  the  axis  for  a  time  of  5  seconds.     Find 
the  speed  and  the  total  number  of  turns,  starting  from 
rest,  made  by  the  wheel  if  it  weighs  64  Ibs.  and  is  of  such 
a  size  and  shape  that  r0  =  9  in. 

3.  A  uniform  circular  disk  4  ft.  in  diameter  weighing 
1000  Ibs.  is  caused  to  rotate  from  rest  by  a  force  of  100  Ibs. 
acting    at    its    circumference.     What    will    be    the    linear 
velocity  of  a  point  on  the  circumference  at  end  of  1J  min- 
utes? 

NOTE.  For  uniform  circular  disk,  axis  at  center,  r0=  approx- 
imately r0  of  radius. 


ELEMENTARY  PRACTICAL  MECHANICS 


171 


4.  What  force  acting  at  the  rim  of  the  disk  of  Prob.  3 
will  be  required  to  change  its  speed  from  300  r.p.m.  to 
120  r.p.m.  in  20  seconds? 

91.  Centrifugal  Force.  —  Because  of  their  inertia,  bodies 
set  in  motion  tend  to  continue  moving  in  a  straight  line. 
Thus  the  body  m,  Fig.  140,  moving  from  the  positions 
MI,  M2,  etc.,  tends  to  go  along  the  straight  paths,  MiNi, 
M2N2,  etc.  To  pull  it  out 
of  a  straight  path,  and  com- 
pel it  to  move  in  a  circle,  re- 
quires a  force  acting  constantly 
toward  the  center  0.  This 
force  toward  the  center, 
which  is  necessary  to  make 
a  body  rotate  about  this 
center,  is  called  the  centrip- 
etal force.  The  equal  and 
opposite  reaction  to  the  cen- 
tripetal force  is  called  cen- 
trifugal force.  It  is  this 
"  centrifugal  force  "  which 
sometimes  causes  rapidly-rotated  fly-wheels,  emery  wheels, 
etc.,  to  burst. 

It  should  be  noted  that  centrifugal  force  is  not  a  force 
in  the  usual  sense  of  the  word,  but  is  a  reaction  due  to  inertia. 

It  can  be  shown  (see  foot-note  page  172)  that  the  force 
that  must  act  toward  the  center  to  keep  a  body  moving  at 
a  uniform  velocity  in  a  circular  path  depends  upon  the  mass 
of  the  body,  upon  the  square  of  the  velocity,  and  upon  the 
size  of  the  circle;  and  that  the  relationship  between  these 
quantities  is  expressed  by  the  equation, 


FIG.  140. 


Centripetal  force=  --  (absolute  units). 
r 


172  ELEMENTARY  PRACTICAL  MECHANICS 

In  the  gravitation  system  which  we   commonly  use,  this 
equation  becomes 


gr 

where  W  is  the  weight  of  the  body,  V  its  velocity  expressed 
in  linear  distance  traveled  per  second,  and  r  the  radius 
of  the  circular  path.  Since  "  centrifugal  force  "  is  equal 
and  opposite  to  centripetal  force,  we  may  also  write  the 
equation, 

mv2 

Centrifugal  torce=  --  = 


gr 

The  rotating  parts  of  all  machinery  must  be  carefully 
balanced  about  the  axis,  otherwise  centrifugal  force  may 
cause  injurious  wear  at  bearings,  jarring,  etc. 

Example.  —  What  is  the  centrifugal  force  of  each  pound 
of  weight  of  the  rim  of  a  fly-wheel  6  ft.  diameter,  making 
400  revolutions  per  minute? 

6X7TX400 
Velocity  =  --  —  --  =  126  ft/  sec  approx. 

1  X  126  X  126 

Therefore  centrifugal  force  =  -      -  =  164  Ibs. 

32.2X3 


NOTE. — Suppose  a  body  of  mass  m  to  be  moving  with  uniform 
velocity  v  around  the  circle  ACE,  Fig.  141.  Let  t  be  the  time  re- 
quired to  move  over  the  arc  AC.  In  the  same  time,  the  body  has 
been  pulled  out  of  its  straight  path  an  amount  equal  to  the  distance 
BC  by  a  force  which,  as  the  path  is  circular,  must  everywhere  be 
directed  toward  the  center  0.  If  a  be  the  acceleration  which  this 

at2 
force  will  impart  to  the  mass  .n,  BC  =  — •.     If  the  time  t  be  sufficiently 

Zi 

short,  arc  AC  will  be  so  short  that  it  may  be  regarded  as  a  straight 


ELEMENTARY  PRACTICAL  MECHANICS 


173 


PROBLEMS 

1.  In  a  machine  running  at  1800  revolutions  per  minute 
there  is  an  unbalanced  weight  of  one  pound  at  a  radius  of 
one  foot.     Find  pull  on  the  bearings  due  to  centrifugal 
force. 

2.  A  bicycle  and  rider  weighing  together  200  Ibs.   run 
around  a  curve  of  radius  50  ft.  at  a  rate  of  12  miles  per 
hour.     Find  force  with  which  tire  must  stick  to  track  to 
prevent  slipping. 

3.  How  many  revolutions  per  minute  will  suffice  to  break 
a  string  of  tensile  strength  of  10  Ibs.  if  a  body  weighing  5  Ibs. 
be  swung  round  at  the  end  of  6  ft.  of  the  string? 

4.  A  train  of  200  tons  weight  is  rounding  a  curve  of 
\  mile  radius  with  a  velocity  of  30  miles  per  hour.     What 
is  the  horizontal  pressure  on  the  rails? 

line:  figures  ACD  and  ACE  will  then  be  similar  triangles.     Therefore, 

~AC  =  ~AE 

(2)        (AC)2  =  ADXAE. 
But          AC=vt  (distance  moved  in  uniform  motion); 


And 


at2 
BC= —  (distance  moved,  accelerated  motion). 

£ 

A  B 


AE  =  2r,  r  being  the  radius  of  the  circle. 
Substituting  these  values  in  equation 

(2)    and   solving,  we  have  a  =  — ,  or  the 

r 

acceleration  imparted  by  the  force  acting 
toward  the  center  of  the  circle  equals  the 
square  of  the  velocity  divided  by  the 
radius  of  the  path.  And  since  force 
equals  mass  multiplied  by  acceleration, 
we  have 

Centripetal  force  =  - 


FIG.  141. 


174  ELEMENTARY  PRACTICAL  MECHANICS 

5.  What  should  be  the  elevation  of  the  outer  rail  for  a 
standard  gauge  track  of  4  ft.  lOf  in.  for  the  train  of  Prob.  4? 

6.  Find  the  speed  at  which  a  simple  Watt  governor  runs 
when  the  arm  makes  an  angle  of  30°  with  the  vertical. 
Length  of  arm  from  center  of  pin  to  center  of  ball  18  in. 

7.  The  rim  of  a  cast-iron  fly-wheel  has  a  mean  radius 
of  2  ft.     Its  section  is  6  in.  broad  and  4  in.  thick.     It  is 
revolving  at  180  r.p.m.     What  is  the  centrifugal  force  per 
inch  length  of  rim? 


CHAPTER  X 
WORK,   POWER,   ENERGY 

92.  Work. — When  a  force  causes  a  body  to  change  its 
position  against  a  resistance  opposing  such  change,  work 
is  done  on  the  body.  Thus  work  is  done  upon  a  body 
when  the  body  is  lifted  against  the  force  of  gravity;  when 
the  body  is  moved  against  the  resistance  offered  by  fric- 
tion; or  when  the  velocity  of  the  body  is  increased  or 
decreased  against  the  reaction  offered  by  inertia. 

Two  items  must  always  be  present  when  work  is  done; 
viz.,  force  (equal  to  the  opposing  resistance),  and  displace- 
ment. The  measure  of  the  work  done  is  the  product  of  the 
force  and  the  displacement  in  the  direction  of  the  force.  Or, 

Work  =  Fs. 

Where  the  displacement  is  not  in  the  une  of  action  of 
the  force,  either  the  component  of  the  displacement  in  the 
direction  of  the  force  must  be  found,  or  the  component  of 
the  force  in  the  direction  of  the  displacement.  The  work 
is  then  equal  to  the  product  of  the  force  and  the  component 
of  the  displacement  in  the  direction  of  the  force;  or,  it  is 
equal  to  the  product  of  the  displacement  by  the  component  of 
the  force  in  the  direction  of  the  displacement.  Thus,  suppose 
a  body  weighing  100  Ibs.  is  to  be  pushed  10  ft.  up  a  smooth 
plane  rising  30°  to  the  horizontal.  The  resistance  here 
is  due  to  the  vertical  gravity  pull  of  100  Ibs.,  the  displace- 
ment in  the  direction  of  this  force  is  10  sin  30°  or  5  ft. 

175 


176  ELEMENTARY  PRACTICAL  MECHANICS 

Hence  the  work  done  is  100X5  =  500  ft.-lbs.  Or,  the 
component  of  the  force  in  the  direction  of  displacement 
is  100  sin  30°  =  50  Ibs.,  and  the  work  done  is  therefore 
50X10  =  500  ft.-lbs.  Or,  suppose  a  body  weighing  200  Ibs. 
resting  upon  a  horizontal  table  is  pushed  3  ft.  along  the 
table  by  a  force  of  60  Ibs.  acting  at  an  angle  of  20°  with 
the  horizontal.  The  force  in  the  direction  of  displacement 
is  here  60  X  cos  20°  =  56.4  Ibs.  The  work  is  therefore 
56.4X3  =  169.2  ft. -Ibs.  The  weight  of  the  body  and  the 
component  of  the  force  perpendicular  to  the  table  do  no 
work  as  they  produce  no  displacement  in  their  directions. 

93.  Units  of  Work. — Quantity  of  work  done  is  measured 
by  the  product  force  X  distance.     Work  may  therefore  be 
expressed   in    different    units,    dependent   upon   the   units 
used  in  defining  the  force  and  the  displacement.     Thus  if 
force  is  measured  in  pounds  and  displacement  in  inches, 
the  work  will  be  expressed  in  inch-pounds.     Other  units 
are  centimeter-grams,  foot-tons,  etc. 

The  units  of  work  commonly  used  are  the  foot-pound 
and  the  erg. 

A  FOOT-POUND  is  the  work  done  when  a  force  of  one  pound 
acts  through  a  distance  of  one  foot  in  the  direction  of  the 
force;  or  when  an  equivalent  amount  of  work  is  done,  as 
when  a  force  of  2  Ibs.  acts  through  a  distance  of  J  ft.,  a 
quarter  pound  acts  through  a  distance  of  4  ft.,  etc. 

AN  ERG  is  the  work  done  by  a  force  of  one  dyne  acting 
through  a  distance  of  one  centimeter.  The  erg  is  thus  a 
very  small  amount  of  work;  for  convenience  of  expression, 
therefore,  the  JOULE  =  107  ergs  is  often  used. 

94.  Work  Diagrams. — Since  work  equals  the  product  of 
force  and   distance,  it  may  often  be  conveniently  repre- 
sented by  the  area  of  a  diagram.     Thus,  suppose  a  con- 
stant force  of  100  Ibs.  acts  through  a  distance  of  16  in.; 
we  may  represent  the  force  by  a  vertical  line  5  units  long 


ELEMENTARY  PRACTICAL  MECHANICS 


177 


FIG.  142. 


and  the  distance  by  a  horizontal  line  4  units  long,  then  the 
area  of  the  figure  represents  the  work  in  terms  of  the  scale 
used.  A  unit  on  the  vertical  scale 
here  represents  20  Ibs.,  one  unit  on 
the  horizontal  scale  represents  4  in. 
of  distance.  One  square  unit,  there- 
fore, represents  20^^  =  80  inch-pounds 
of  work,  and  the  whole  area  5X4  = 
20  sq.  units  represents  20X80  =  1600 
in.-lbs. 

If  the  force  doing  work  is  not 
constant,  the  work  done  may  still 
be  represented  by  an  area.  Thus, 
suppose  Fig.  143  represents  the  con- 
ditions in  a  steam-engine  cylinder  where,  as  is  usually 
the  case,  the  steam  pressure  is  not  constant  during  the 
entire  stroke.  The  vertical  height  of  the  diagram  at  any 
point  represents  the  steam  pressure  in  pounds  per 
square  inch  at  that  point  in  the  stroke.  The  area  of 
the  figure  (  =  average  height  X  length)  therefore  represents, 
in  terms  of  the  scale  to 
which  it  is  drawn  the  prod- 
uct of  average  steam  pres- 
sure times  length  of  stroke 
or  the  work  done  by  the 
steam  on  each  square  inch 
of  the  piston  as  it  moves 
forward  the  length  of  one 
stroke. 

Suppose  the  area  of  the 

diagram  is  2.2  sq.  in.  arid  suppose  TO  in.  =  1  in.  is  the 
horizontal  scale,  and  1  in.  =  30  Ibs.  per  square  inch  is 
the  vertical  scale.  Then  each  square  inch  of  the  dia- 
gram represents  10x30  =  300  in.-lbs.,  and  the  whole 


FIG.  143. 


178  ELEMENTARY  PRACTICAL  MECHANICS 

diagram  represents  2.2X300  =  660  inch-pounds  or  55  foot- 
pounds. 

The  area  of  irregular  figures  such  as  the  above  may  be 
found  approximately  by  measuring  the  height  at  10  or 
more  points  at  equal  intervals  along  its  entire  length  and 
taking  the  average  of  these  measured  values  as  the  average 
height.  Average  height  X  length  =  area.  A  more  accurate 
method  is  to  use  an  instrument  for  measuring  areas  known 
as  a  planimeter. 

Work  diagrams  similar  to  Fig.  143,  taken  from  engines  in 
actual  practice,  are  known  as  indicator  cards;  instruments 
for  taking  indicator  cards  are  known  as  steam-engine  indi- 
cators. In  practice,  steam-engine  indicators  are  "  cali- 
brated "  so  that  the  values  of  the  vertical  and  horizontal 
scales  of  the  diagram  in  terms  of  actual  steam  pressure 
and  length  of  stroke  are  known. 

95.  Power. — In  the  previous  discussion  of  work  it  will 
be  noticed  that  the  time  required  for  the  force  to  act  through 
a  given  distance  was  not  considered;    the  total  work  being 
the  same  whether  performed   quickly   or  slowly.     But   a 
machine  which  will  do  a  given  amount  of  work  in  a  short 
time  is  said  to  be  "  more  powerful  "  than  one  that  requires 
greater  time  in  which  to  do  the  same  work. 

POWER  is  therefore  defined  as  the  rate  of  doing  work. 

96.  Units  of  Power. — Power  may  be  expressed  in  any 
units  which  specify  the  work  done  and  the  time  in  which 
it  is  performed.     The  customary  units  of  power  are,  how- 
ever, the  horse-power  and  the  watt. 

Work  is  done  at  the  rate  of  ONE  HORSE-POWER  (abbre- 
viated H.P.)  when  33,000  foot-pounds  of  work  are  done  in 
one  minute.  Therefore, 

foot-pounds  in  one  minute 

"  '"  33,000  :•;' 


ELEMENTARY  PRACTICAL  MECHANICS  179 

Work  is  done  at  the  rate  of  ONE  WATT  when  one  joule  of 
work  is  done  per  second. 

1  H.P.=  746  watts. 

1  kilowatt  =  1000  watts  = ,  or  about  H  H.P. 

746 

In  measuring  electrical  power  the  product  of  the  elec- 
trical pressure  by  the  current  in  amperes  which  it  is  pro- 
ducing =  watts.  Or  watts  =  volts  X  amperes. 


PROBLEMS 

1.  A  car  weighing  3000  Ibs.  and  containing  2  tons  of 
coal  is  hoisted  up  the  shaft  of  a  mine  400  ft.  deep.     Work 
done? 

2.  A  magnet  is  moved  124  cm.  against  a  resistance  of 
1200  dynes.     Work  done? 

3.  Weight  of  a  trolley  car  is  8  tons.     Average  resistance 
to   motion   due  to   friction,   etc.,   is   one-sixteenth   of   the 
weight.     Work  done  in  run  of  4  miles  on  a  level  track? 
Power  required  if  run  is  made  in  20  minutes? 

4.  James  Watt  found  that  an  ordinary  English  dray-horse 
could  travel    an  average  of  2^  miles  per  hour  and    raise 
150  Ibs.  by  means  of  a  rope  and  pulley.     Find  the  work 
done  per  minute. 

5.  A  man  weighing  180  Ibs.  walks  up  stairs,  a  height  of 
20  ft.  in  J  minute.     At  what  rate  does  he  work? 

6.  A  load  is   hauled  along  a  level  floor  by  means  of  a 
rope   inclined   30°  to  the   horizontal.     If  pull  in   rope  is 
100  Ibs.  and  the  body  is  moved  10  ft.,  compute  work  done. 

7.  Find   H.P.   of  the   engine   that   should   be   used   for 
raising  coal  from  a  pit  200  ft.  deep,  working  15  hours  a 
day,  if  the  average  daily  yield  is  1000  long  tons. 


180  ELEMENTARY  PRACTICAL  MECHANICS 

8.  A  train  weighs  220  tons.     The  resistance  to  motion  is 
20  Ibs.  per  ton  on  a  level.     Find  H.P.  of  an  engine  which 
can  just  keep  it  going  30  miles  an  hour  up  a  grade  of  10  ft. 
rise  per  mile  of  track. 

9.  A  5-H.P.  engine  is  used  to  pump  water  from  a  shaft 
100  ft.   deep.     How  many  cubic  feet  will  it  raise  in  24 
hours?     (Cubic  foot  of  water  weighs  62.5  Ibs.) 

10.  What  H.P.  is  required  to  supply  1000  families  with 
60  gals,  water  each  per  day  of  10  hours,  if  water  is  pumped 
to  a  height  of  200  ft.?     (Gallon  of  water  weighs  8  Ibs.) 

11.  Find  the  work  done  against  a  brake  which  gives  a 
frictional  resistance  of   15  Ibs.   on  the  rim  of  a  wheel  of 
2  ft.  radius,  while  the  wheel  makes  50  revolutions. 

12.  Find  the  H.P.  used  in  the  preceding  problem  if  the 
50  revolutions  required  20  seconds  time. 

13.  Compute  the   H.P.   developed  by  an  engine  under 
the  following  conditions:    Diameter  of  piston  4  in.;    length 
of  stroke  5  in. ;    mean  effective  pressure  42  Ibs.  per  square 
inch;  number  of  revolutions  275  per  minute. 

14.  A    body    4  ft.X3  ft.X2  ft.    rests    upon    2X3    face. 
Body  weighs  150  Ibs.     Compute  the  work  done  in  turning 
it  over  on  2  ft.  edge. 

15.  An  engine  hauls  a  car   10,000  ft.   along  the  track. 
The  pull  exerted  by  the  engine  upon  the  car  in  the  begin- 
ning was  3500  Ibs.  and  as  recorded  every  1000  ft.  was  as  fol- 
lows:   3500,    3250,    2800,    3450,    3700,    3600,    3550,    3300, 
2950,   and  2750  Ibs.     Construct  a   "  work   diagram  "    for 
above  data.     Determine  area  of    diagram   and    from    this 
the  total  work  done  on  car. 

16.  A  punch  exerts  an  average  pressure  of  50  tons  in 
punching  a  hole  through  a  plate   H  in.  thick.     Compute 
work  done. 

17.  A  well  6  ft.  diameter  and  30  ft.  deep  is  dug.     Com- 
pute work  'done  in  raising  the  material  if  1  cu.  yd.  weighs 
4000  Ibs. 

18.  The  weight  of  a  train  is  240  tons  and  the  draw-bar 
pull  is  8  Ibs.  per  ton.     Find  H.P.  required  to  keep  train 
running  20  miles  per  hour. 


i  ELEMENTARY  PRACTICAL  MECHANICS  181 

19.  What  is  the  difference  in  tension  between  the  two 
sides  of  a  belt  running  3000  ft.  a  minute  and  transmitting 
250  H.P.? 

20.  Find  the  speed  of  a  driving  pulley  2J  ft.  diameter  to 
transmit  10  H.P.,  the  driving  force  of  the  belt  being  120  Ibs. 

21.  An  automobile  that  weighs  2  tons  goes  up  a  rough 
road  of  1  ft.  rise  in  10  ft.  of  road  at  a  speed  of  15  miles 
per  hour.     Frictional  resistance  is  18  Ibs.  per  ton.     Com- 
pute H.P.  developed. 

22.  Average  width  of  an  indicator  diagram  for  one  end 
of  a  piston  is  1.6  in.     For  the  other  1.5  in.,  and  1  in.  repre- 
sents 36  Ibs.  per  square  inch.     Piston  is  10  in.  diameter, 
stroke    1   ft.,   revolutions   per   minute    120.     What   is  the 
indicated  H.P.  if  both  sides  of  the  piston  are  equal? 

23.  What  is  the  H.P.  of  a  stream  that  has  a  section  of 
10  sq.  ft.,  velocity  of  3  miles  per  hour,  and  a  12-ft.  fall? 

24.  A  fire-pump  will  deliver  1000  gals,  of  water  per  minute 
at  100  Ibs.  pressure.     Compute  \york  expended  in  pumping. 
(1  Ib.  pressure  =  2. 34  ft.-head.) 

97.  Energy. — A  body  which  is  capable  of  doing  work  is 
said  to  possess  energy. 

Thus,  we  say  that  a  clock  spring  when  coiled  possesses 
energy  because,  in  unwinding,  it  can  do  work  in  driving 
the  clock  mechanism,  that  a  moving  projectile  possesses 
energy  because  it  can  overcome  the  resistance  offered  by 
the  air,  by  armor  plate,  etc.,  and  thus  do  work.  A  storage- 
battery  possesses  energy  because  it  can  furnish  an  electric 
current  to  operate  a  motor,  etc.  Hence, 

Definition:  ENERGY  is  THE  CAPACITY  FOR  DOING  WORK. 

98.  Source  of  the  Energy  of  a  Body. — A  body  or  system 
of  bodies  possess  energy  as  a  result  of  the  fact  that  at  some 
previous  time  work  has  been  done  upon  it.     Thus   in  the 
illustrations  just  given,  the  coiled  spring  received  its  energy 
when  work  was  done  on  it  in  winding  it  up;   the  moving 


182  ELEMENTARY  PRACTICAL  MECHANICS 

projectile  acquired  the  energy  it  did  not  have  while  lying 
at  rest  in  the  cannon  when  the  gases  produced  in  the 
explosion  of  the  powder,  by  their  presure,  did  work  upon 
the  projectile,  thus  overcoming  the  resistance  offered  by 
its  inertia  and  imparting  motion;  the  storage-battery 
received  its  energy  when  "  charged  "  from  the  dynamo 
machine.  Work  is  therefore  the  act  of  transferring  energy. 
The  body  which  is  doing  the  work  gives  or  expends  energy, 
the  body  worked  upon  receives  energy. 

The  energy  thus  transferred  may  be  transformed  into 
heat  during  the  process,  and  thus  be  no  longer  available 
for  doing  wTork;  or  a  part  of  it  may  be  stored  up  as  avail- 
able energy  in  the  body  worked  upon,  as  when  this  body  is 
raised  to  a  higher  level,  when  it  is  put  in  motion,  etc. 
When  energy  is  thus  stored,  the  body  worked  upon  becomes 
in  turn  capable  of  doing  work. 

The  amount  of  available  energy  possessed  by  a  body  is 
measured  by  the  amount  of  work  the  body  can  do  and  is 
therefore  expressed  in  the  same  units  as  work,  i.e.,  in 
foot-pounds,  ergs,  etc. 

99.  Kinetic  and  Potential  Energy. — Mechanical  energy  is 
of  two  kinds: 

First,  Energy  of  Position,  or  Potential  Energy.  Thus,  if 
a  compressed  spring  can  in  extending  exert  a  force  of 
60  Ibs.  over  a  distance  of  4  in.,  it  has  when  compressed  a 
potential  energy  of  240  in.-lbs.;  or,  if  a  body  weighing 
10  Ibs.  is  suspended  so  that  when  released  it  may  freely 
fall  under  the  action  of  gravity  6  ft.,  it  has  a  potential 
energy  due  to  its  position  of  60  ft.-lbs. 

Second,  Energy  due  to  Motion,  called  Kinetic  Energy.  A 
body  in  motion  can  be  brought  to  rest  only  through  a 
resistance  opposing  the  motion.  The  product  of  this 
resistance  by  the  distance  through  which  it  acts  in  stopping 
the  body  measures  the  work  stored  in  the  body,  and  there- 


ELEMENTARY  PRACTICAL  MECHANICS  183 

fore  the  kinetic  energy  which  it  possesses.  Thus  the 
kinetic  energy  of  a  moving  train  =  the  work  done  by  it 
while  stopping  =  the  resistance  due  to  friction,  brakes,  etc., 
X  distance  moved  over  while  coming  to  rest. 

The  kinetic  energy  of  a  body  may  be  computed  from  its 
mass  and  velocity  as  follows: 

Kinetic  energy  =  work  a  body  can  do  because  of  its 
motion  =  retarding  force  X  distance  through  which  it  acts 
in  stopping  body  =  FXs.  But 

F  =  ma. 
Therefore,  K.  E.  =  Fs  =  mas. 

v  at2 

When  a  body  is  accelerated  or  retarded,  a=—  and  s  =  —  , 

therefore,  by  substituting  these  values  in  the  equation 
above,  we  have, 


Wv2 
In  gravitational  units,  this  expression  becomes:  K.E.  =  -  . 

<y 

In  applying  this  equation,  v  and  g  must  always  be  ex- 
pressed in  the  same  units. 

100.  Transformation  of  Energy.  —  Potential  energy  may 
be  transformed  into  an  equal  amount  of  kinetic  energy, 
and  vice  versa,  the  sum  of  the  two  kinds  of  energy  always 
remaining  constant  unless  some  of  it  is  used  up  in  over- 
coming friction,  or  in  doing  other  work.  Thus,  if  a  body 
has  potential  energy,  due  to  its  position,  and  is  allowed  to 
fall,  its  potential  energy  decreases  as  it  descends.  But  its 
kinetic  energy  due  to  its  velocity,  increases  at  the  same 
rate,  until  the  body  has,  just  as  it  is  reaching  the  end 
of  its  fall,  no  potential  energy,  but  an  amount  of  kinetic 
energy  equal  to  the  potential  energy  which  it  had  at  the 


184  ELEMENTARY  PRACTICAL  MECHANICS 

start.     If  a  body  weighing  W  Ibs.  is  suspended  h  ft.  above 
the  ground  its  potential  energy  is  Wh  ft. -Ibs.     If  it  falls 

Wv2 
freely  its  kinetic  energy  on  striking  the  ground  =• =  Wh. 

o 

Similarly,  if  a  body  has  a  certain  amount  of  kinetic 
energy,  due  to  its  velocity,  and  if  this  velocity  is  directed 
upward,  the  velocity  of  the  body,  and  therefore  its  kinetic 
energy,  will  decrease  as  it  rises;  but  its  potential  energy 
will  increase  as  it  rises  in  the  same  ratio  until,  when  its 
velocity  has  been  reduced  to  zero  and  it  is  just  ready  to 
commence  to  fall  again,  its  potential  energy  will  exactly 
equal  the  kinetic  energy  which  the  body  had  in  the  beginning. 

In  addition  to  what  has  here  been  termed  "  mechanical 
energy/7  energy  may  exist  in  other  forms,  as  heat,  elec- 
trical energy,  chemical  energy,  etc.  Under  proper  con- 
ditions, energy  of  one  form  may  be  transformed  into 
another  form.  Thus  mechanical  energy  may  be  trans- 
formed into  the  form  of  molecular  kinetic  energy  called 
heat,  as,  for  example,  in  machinery  where  energy  expended 
against  friction  is  transformed  into  heat  thus  heating  the 
bearings.  In  improperly  constructed  or  poorly  lubricated 
bearings  the  amount  of  energy  thus  transformed,  i.e.,  the 
"wasted  work,"  may  be  considerable,  and  thus  the  parts 
of  the  machine  may  be  injured  from  undue  heating. 

Or,  heat  energy  may  be  transformed  into  mechanical 
energy,  as  in  the  steam-engine. 

Whenever  heat  energy  is  transformed  into  mechanical 
energy,  or  mechanical  energy  into  heat,  it  has  been  defi- 
nitely determined  that  the  transformation  always  occurs 
in  accordance  with  the  law  that : 

The  heat  energy  required  to  heat  1  Ib.  of  water  1°  F.  =  778 
ft.-lbs.  (See  also  Art.  112.) 

Mechanical  energy  when  used  for  driving  a  dynamo  may 
be  transformed  into  the  energy  of  an  electric  current,  and 


ELEMENTARY  PRACTICAL  MECHANICS  185 

by  means  of  a  motor,  energy  of  an  electric  current  is 
transformed  into  mechanical  energy.  In  electric  lights, 
heaters,  etc.,  electrical  erergy  is  transformed  into  heat. 


PROBLEMS 

1.  A  reservoir  contains  50,000  gals,  water  at  an  average 
elevation  200  ft.  above  a  given  level.     Potential  energy  of 
watei  with  respect  to  this  level? 

2.  What  is  the  potential  energy  of  a  mass  of  1500  Abs. 
at  a  height  of  80  ft.  above  the  ground?     What  will  be  its 
velocity  on  reaching  the  ground  if  allowed  to  fall  freely 
and  what  will  be  its  kinetic  energy  on  striking? 

3.  A  bullet  weighing  15  gms.  has  a  muzzle  velocity  of 
600  meters  per  second.     How  far  would  the  energy  of  the 
bullet  raise  a  mass  of  1  kgm.  if  it  could  all  be  used  for  that 
purpose? 

4.  A  5-ton  pile-driver  falls  freely  6  ft.     Assuming  all  of 
its    kinetic   energy   to   be   expended   in   sinking   the   pile, 
against  what  average  resistance  can  it  drive  the  pile  1  in.? 

5.  If  the  mass  of  a  moving  body  be  doubled  and  its 
velocity  at  the  same  time  be  increased  fourfold,  how  is 
the  kinetic  energy  changed? 

6.  6480  ergs  of  energy  are  expended  by  the  action  of  a 
constant  force  urging  a  mass  of  90  gms.  a  distance  of  600  cm. 
along  a  smooth  horizontal  plane.     Find 

(a)  The  final  velocity; 

(b)  The  acceleration; 

(c)  The  time; 

(d)  The  force. 

7.  A  car  weighing  10  tons  is  moving  at  the  rate  of  36 
miles  an  hour.     Find   average  force   required  to   stop  it 
within  a  space  of  50  ft. 

8.  A  bicyclist  weighing   155  Ibs.   mounted  on  a  wheel 
weighing  25  Ibs.  rides  at  a  velocity  of  25  miles  per  hour 
on  a  horizontal  plane.     If  he  comes  to  a  slope  rising  12° 


186 


ELEMENTARY  PRACTICAL  MECHANICS 


and  removes  his  feet  from  pedals,  how  far  up  the  slope 
will  he  go? 

9.  The  steam  hammer  shown  in  the  diagram  (Fig.  144) 

has  a  cylinder  72  sq.in.  ir?  cross-section, 
the  stroke  is  20  in.,  the  steam  pressure 
is  80  Ibs.  per  square  inch,  the  weight 
of  hammer,  ir  eluding  piston  and  piston- 
rod,  ie  644  Ibs.  Find  the  energy  which 
hammer  will  have  after  one  full  stroke. 
How  far  will  the  hammer  blow  sink 
into  a  piece  of  copper  if  the  average 
resistance  offered  by  the  copper  is  50 

tons? 
FIG.  144. 

NOTE.     Work  done  by  gravity  and  work 

done  by  steam,  both  give  velocity  to  hammer.     Disregard  friction. 

10.  The  section  of  a  stream  is   10  square  ft.   and  the 
average  velocity  of  the  water  2.  ft /sec;   there  is  an  avail- 
able fall  of  30  ft.     What  is  the  available  supply  of  energy 
per  minute? 

11.  A  ball  weighing  ^  Ib.  strikes  a  shield  with  a  velocity 
of  1000  ft /sec.     It  pierces  the  shield  and  moves  on  with  a 
velocity   of   400   ft /sec.     Energy  lost   in   piercing   shield? 
Resistance  offered  by  shield  if  it  is  1  in.  thick? 

12.  A  fire  engine  pump  is  provided  with  a  nozzle  1  sq.in. 
area  of  section  and  water  is  projected  through  the  nozzle 
with  an  average  velocity  of  140  ft /sec.     Find: 

1.  Kinetic  energy  of  each  pound  of  water  as  it  leaves 

nozzle. 

2.  Horse-power  engine  required  to  drive  the  pump 

which  supplies  the  nozzle,  assuming  that  75  per 
cent  of  the  energy  supplied  to  the  pump  is 
transferred  to  the  water. 

13.  A  rapid-fire  gun  discharges  5  projectiles  per  minute 
each  of  a  weight  of  100  Ibs.  and  a  velocity  of  2000  ft/sec. 
What  is  the  H.P.  expended? 

14.  A  hammer  weighing  1J  Ibs.  has  a  velocity  of  18  ft/sec, 
when  it  strikes  a  nail.     What  is  the  force  exerted  on  the 
nail  if  it  is  driven  into  the  wood  i  inch? 


ELEMENTARY  PRACTICAL  MECHANICS  187 

101.  Energy  of  Rotating  Bodies.  —  As  shown  in  Art.  99, 
the  kinetic  energy  of  a  body  moving  in  a  straight  line  is 
expressed  by  half  the  product  of  its  mass  by  the  square 
of  its  velocity.  In  such  motion  all  portions  of  the  body 
move  in  the  same  direction  and  with  the  same  velocity. 
In  the  case  of  a  rotating  body,  as  for  example,  a  fly-wheel, 
the  kinetic  energy  of  any  particular  particle  of  the  wheel 
is  obviously  %mv2,  where  m  is  the  mass  of  the  particle,  v 
its  velocity,  and  the  kinetic  energy  of  the  whole  wheel  is 
the  sum  of  the  kinetic  energy  of  its  particles,  or  sum  (%mv2), 
where  v  has  different  values  according  to  the  distance  of 
the  particles  from  the  axis  of  rotation. 

Since  in  rotary  motion  the  whole  mass  of  the  moving 
body  may  be  assumed  as  concentrated  at  a  distance  equal 
to  the  radius  of  gyration  from  the  axis  (see  Art.  89),  a 
convenient  equivalent  expression  for  the  kinetic  energy  of 
a  rotating  body  is 

Wv  2 


where  ^o^the  linear  velocity  at  the  distance  of  the  radius 
of  gyration  from  the  axis.* 

Example  1.  —  What  is  the  kinetic  energy  of  a  fly-wheel 
weighing  3000  Ibs.  when  running  180  revolutions  per  minute, 
its  dimensions  being  as  follows  : 

Outside  diameter  6  'ft.  6  in.; 

Thickness  of  rim  6  in.; 

Spokes  and  hub  may  be  neglected. 

*  Since  moment  of  inertia  bears  the  same  relation  to  rotary  motion 
as  mass  does  to  linear  motion,  the  kinetic  energy  of  a  rotating  body 
may  also  be  expressed  as  one-half  the  product  of  its  moment  of 
inertia  by  the  square  of  its  angular  velocity,  or, 


188  ELEMENTARY  PRACTICAL  MECHANICS 

The  radius  of  gyration  for  fly-wheels  where  spokes  and 
hub  may  be  neglected  is  approximately  equal  to  the  mean 
radius  of  the  rim.  Hence  the  kinetic  energy  in  this  case 
will  be  approximately  as  if  all  mass  were  placed  3  ft.  from 
the  center.  Therefore, 

6X5X180 

v=,  -  =  56.5  ft/sec.; 

K.  E.==30°5  =  149,000  ft,lbs. 


Example  2.  —  A  fly-wheel  weighing  5400  Ibs.,  mean  radius 
2  ft.  9  in.  running  at  120  turns  per  minute,  gives  up  8000 
ft.-lbs.  of  energy;  how  much  is  its  velocity  diminished? 

Let  v\  be  velocity  in  beginning,  and 

V2  be  velocity  after  the  8000  ft.-lbs.  have  been  with- 
drawn. 

2X2fX7rXl20 

vi  =  ---  —  -  =34.6  ft  /sec; 
50 

5400  X^i2 


K.  E.  in  beginning  = 


K.  E.  at  end  = 


2X32.2  ' 
5400  Xv22 


2X32.2 
5400  X^i2     5400  X^22 


_, 
Therefore, 

Substituting  the  value  of  v\  above  arid  solving, 
#2  =  32.2  ft /sec  =  115  + revolutions  per  minute. 


.     ELEMENTARY  PRACTICAL  MECHANICS  189 

102.  Experimental  Study  of  Kinetic  Energy  of  a  Fly- 
wheel.—^The  student  apparatus  shown  in  Fig.  116  may 
be  used  to  illustrate  'experimentally,  the  law  of  conserva- 
tion of  energy,  the  stored  kinetic  energy  of  a  rotating 
body,  and  such  terms  as  radius  of  gyration,  moment  of 
inertia,  etc.,  which  are  apt  to  be  hazy  and  very  indefinite 
conceptions.  For  this  purpose  the  cord  leading  from  the 
scale  pan  ends  in  a  loop  passed  over  a  pin  in  the  drum 
and  the  length  of  cord  is  so  fixed  that  the  loop  is  released 
from  the  pin  at  the  instant  the  weight  reaches  the  floor. 
The  energy  supplied  to  the  system  wrhile  the  weight  descends 
is  represented  by  the  product  Wh,  where  W  is  weight  of  scale 
pan,  h  the  vertical  distance  through  which  jt  moves  before 
it  is  released.  This  energy  must  have  been  expended  in 
work  against  friction  while  the  torque  was  being  applied 
or  be  stored  as  kinetic  energy  in  the  moving  pulley  and 
scale  pan  at  the  instant  the  latter  is  released.  By  the  law 
of  conservation  of  energy,  the  total  energy  received  must 
equal  the  energy  expended  +  energy  thus  stored,  or 

Wh  =  energy  spent  against  friction  +  kinetic  energy 
of  weight  +  kinetic  energy  of  pulley. 

To  check  this  equation  experimentally,  the  distance  h 
may  be  measured,  and  the  friction  of  the  pulley  deter- 
mined in  terms  of  the  pull  tangent  to  the  drum  required 
to  keep  the  pulley  turning  at  uniform  speed.  In  an  actual 
test,  distance  h  was  found  to  be  26.1  in.,  friction  to  be 
10  ounces.* 

Several  determinations  were  then  made  with  a  stop-watch 
of  the  time  required  for  the  pulley  to  come  to  rest  after 

*  Since  the  friction  of  the  apparatus  as  used,  in  the  experiments 
on  acceleration  was  too  small  to  admit  of  satisfactory  measure- 
ment,, the  cone  bearings  were  tightened  slightly  for  this  test. 
They  were  then  kept-  flooded  with  oil. 


190 


ELEMENTARY  PRACTICAL  MECHANICS 


the  torque  had  been  removed,  and  of  the  number  of  revo- 
lutions made  in  the  meantime.  These  agreed  unexpectedly 
well  as  shown  by  the  following  table: 


Observation. 

Time  in  Seconds. 

Revolutions. 

1 

21.0 

42 

2 

20.0 

39 

3 

20.0 

39 

4 

20.0 

40 

5 

21.0 

42 

6 

20.4 

41 

7 

21.0 

42 

8 

20.0 

40 

The  average  time  required  to  come  to  rest  was  20.4 
seconds,  and  the  average  number  of  revolutions  41.  (In 
this  as  in  subsequent  computation,  two  significant  figures 
only  are  kept,  as  the  method  did  not  seem  to  warrant 
greater  claim  to  precision.) 

The  average  velocity  of  the  pulley  while  coming  to  rest 

41 
was  therefore -  =  2.0  revolutions  per  second.     And  since 

the  final  velocity  was  zero,  at  the  time  the  weight  was 
released  the  wheel  must  have  been  making  2.0  X  2  or  4.0  revo- 
lutions per  second.  The  diameter  of  the  drum  on  which 
the  cord  is  wound  is  3  in.  Therefore  the  velocity  of  the 

3      22 

weight  when  it  struck  the  floor  was  — X — X4. 0  =  3.1  ft/sec. 

12       7 

The  scale  pan  weighed  10.3  Ibs.  Hence,  substituting 
preceding  values,  we  have 

26  1 
(a)  Energy  supplied  =  Wh  =  10.3  X  — ^  =  22.4  ft.-lbs. 


ELEMENTARY  PRACTICAL  MECHANICS  191 

(6)  Energy  expended  in  doing  work  against  friction  = 
force  of  friction  X  distance  through  which  it  was  overcome  = 
10     26. 1_ 
16X~12~  = 

,,  „  .       .  ,  .     Wv2     10.3X3.1X3.1 

(c)  Kinetic  energy  of  weight  = = =  1.5 

2g  2X32 

ft.-lbs. 

(d)  Stored  energy  of  pulley  =  work  done  against  friction 
in  coming  to  rest  =  friction  measured  at  the  drumXcircum- 

10      3      22 

ference  of  drum  X  revolutions  in  coming  to  rest= — X — X — 

lo     \2t      7 

X 4*  =20  ft.-lbs. 

Therefore,  substituting  these  values  in  our  expression  of 
the  law  of  conservation  of  energy — (a)  should  equal  (6)  + 
(c)  +  (d),— we  have,  22.4  ft.-lbs  should  =  1.4  + 1.5+20  ft.-lbs., 
or  22.4  should  equal  22.9.  This  result,  while  far  from 
precise,  illustrates  the  law  very  satisfactorily. 
:  The  kinetic  energy  of  the  pulley  may  be  expressed  by 

wev2 

,  where  v  is  taken  as  nm  speed,  and  we  is  an  expres- 

2x32 

sion  for  the  £J  equivalent  weight  "  of  the  pulley,  i.e.,  the 
weight  which  concentrated  at  the  rim  and  moving  with 
rim  speed  would  oppose  the  same  rotational  inertia  to 
increase  or  decrease  of  speed,  and  would  possess  the  same 
relation  to  energy  changes  as  the  true  weight  of  the  pulley, 
which  is  of  course  distributed  in  the  hub,  spokes,  rim, 
drum,  etc.,  with  different  actual  linear  speeds  at  each  point. 
Equating  this  with  the  value  for  the  kinetic  energy  as  found 
in  (d),  and  putting  for  v  the  computed  rim  speed  of  a  wheel 
13|  in.  diameter,  making  4  revolutions  per  second,  we  have 

/13f     22       \2 

"•    Vl£xTx4) 

-* --20,  from  which  we  =  6.2  Ibs. 


2X32 


192  ELEMENTARY  PRACTICAL  MECHANICS 

This  differs  of  course  from  the  actual  weight  of  the 
pulley  which  is  13  Ibs. 

Or  if,  instead  of  assuming  the  rim  speed  as  the  velocity 
of  the  whole  pulley  and  computing  its  "equivalent  weight  " 
at  this  speed  as, in  the  preceding,  we  assume  the  actual 
weight  13  Ibs.,  and  figure  the  speed  with  which  this  whole 
weight  must  move  in  order  to  possess  the  same  kinetic 
energy  as  the  actual  pulley,  we  have  the  equation, 

13i>2 

-  =  20,  from  which  v  =  9.9  ft/sec. 
2  X  o  2 

This  is  the  speed  of  a  point  4.7  in.  out  from  the  axis  of 
the  pulley,  or,  in  other  words,  what  is  commonly  called 
the  radius  of  gyration  for  the  pulley  is  4.7  in. 

And  since  "  moment  of  inertia,"  7,  of  pulley  is  equal  to 
its  mass  times  the  radius  of  gyration  squared, 


/  ==  13  X  - 


This  agrees  reasonably  well  with  the  value  1.9  which  was 
obtained  independently  by  a  second  experiment,  using 
the  torsional  pendulum  method. 

Of  course  the  above  values  for  "  equivalent  weight/' 
radius  of  gyration,  and  moment  of  inertia  are  approxi- 
mations to  the  same  degree  that  20  ft.-lbs.  expresses  the 
kinetic  energy  of  the  pulley.  They  cannot  be  computed 
directly,,  however,  because  of  the  different  materials  in  the 
various  parts  of  the  pulley,  its  irregularities  in  shape,  the 
balancing  weights  attached  to  the  rim,  etc.,  and  probably 
these  values  compare  well  with  any  which  could  be  obtained 
experimentally. 


ELEMENTARY  PRACTICAL  MECHANICS  193 


PROBLEMS 

1.  The  fly-wheel  of  a  gas  engine  has  following  dimensions: 

Mean  diameter  of  fly-wheel,  6  ft.; 
Width  of  rim,  8  in.; 
Thickness  of  rim,  6  in.; 
Weight  of  rim  per  cubic  foot,  465  Ibs. 
What  is  the  kinetic  energy  of  the  wheel  when  running  at 
150  revolutions  per  minute? 

2.  How  much  energy  must  be  supplied  to  the  wheel  in 
Problem   1  to  increase  its  speed  to    180   revolutions   per 
minute? 

3.  The  rim   of   a   fly-wheel  weighs   15,000  Ibs.   and  its 
mean  linear  velocity  is  40  ft /sec.     How  many  foot-pounds 
of  work  are  stored  in  it?     If  it  is  required  to  give  out 
20,000  ft.-lbs.,  how  much  will  its  velocity  be  decreased? 

4.  A  fly-wheel  has  a  mean  radius  of  3  ft.  4  in.,  and  a 
normal   speed   of   122   revolutions   per   minute.     It   is   re- 
quired to  supply  3600  ft.-lbs.  from  its  store  of  energy  while 
slowing  down  to  120  revolutions  per  minute.     What  mass 
of  rim  is  required? 

5.  A  fly-wheel  weighing  160  Ibs.  .possesses  800  ft.-lbs.  of 
kinetic  energy.     What  is  the  velocity  of  the  point  where 
all  its  mass  may  be  regarded  as  collected? 

103.  Momentum. — The  product  of  the  mass  of  a  moving 
body  by  its  velocity  is  called  its  Momentum.  Or, 

Momentum  =  mv. 

Thus  a  body  of  10  units  mass  when  it  is  moving  with 
velocity  8  units  per  second,  has  a  momentum  80;  a  body 
whose  mass  is  20  units  has  also  a  momentum  of  80  if  its 
velocity  is  4  units  per  second. 

fi     The  starting  and  stopping,  or  any  change  in  the  velocity 
?•  of  a  moving  body  is  governed  by  the  relation  /=  ma.     If 


194  ELEMENTARY  PRACTICAL  MECHANICS 

v 
v= total    change    in    velocity,    a  — — .     Hence    the    above 

i 

mv 

equation   may  be   expressed   in  the  form  /= — ,  i.e.,  the 

i 

force    acting    may   be    measured   by   the    momentum  it   can 
generate  or  destroy  in  unit  time. 

Two  methods  may  therefore  be  used  in  computing  the 
force  acting  in  producing  a  change  of  velocity.  Thus, 
suppose  we  are  given~the  mass  and  velocity  of  the  body 
and  the  distance  through  which  the  force  acts.  We  may 
then  compute  the  kinetic  energy  of  the  body  and  apply 
the  principle  of  work.  For  example,  suppose  a  hammer 
weighing  5  Ibs.,  moving  with  a  velocity  of  40  ft/ sec  when 
it  strikes  a  nail,  drives  the  nail  1  in.  into  the  wood.  The 
kinetic  energy  of  the  hammer  is 

5X40X40 

—  =  124  ft.-lbs. 
2X32.2 

This  energy  is  expended  in  doing  work  against  the  resist- 
ance R  offered  by  the  wood,  or  1 24  =  RX^}  hence, 

R  =  1490  Ibs. 

It  should  be  noted  that  R  is  here  the  average  resistance 
offered  by  the  wood,  over  the  distance  of  1  in.  R  is  the 
space  average  of  the  force.  The  resistance  offered  by  the 
wood  may  or  may  not  be  uniform;  it  is  equivalent  to  the 
uniform  resistance  of  1490  Ibs.  through  the  whole  distance, 
1  inch. 

Or,  if  we  know  the  ?nass  and  velocity  of  the  body  and 
the  time  during  which  the  force  acts,  we  may  compute 
the  momentum  of  the  body  and  then  find  the  force  from 


ELEMENTARY  PRACTICAL  MECHANICS  195 

the  relation  /= — .    Thus,   suppose  the  hammer  of  the 
• 

preceding  illustration  is  brought  to  rest  in  i^th  of  a  second. 


Then   momentum   of  hammer  =  -  X40.     And  the  re- 

32.2 

Wv        5  X  40 
sistance  R  offered  by  the  wood=  -  =  --  =  12401bs. 


. 

R  is  here  the  time-average  force,  i.e.,  the  average  force 
which  will  stop  the  hammer  in  the  given  time,  -%fa  second. 

This  method  is  very  useful  when  considering  impulsive 
forces  or  blows,  i.e.,  forces  which  act  for  a  very  short 
interval  of  time.  The  following  examples  will  illustrate 
the  general  applications: 

Example  1.  —  A  stream  of  water  2  in.  diameter  impinges 
at  an  angle  of  90°  upon  a  brick  wall.  If  the  velocity  of 
the  stream  is  200  ft/  sec,  what  is  the  force  exerted  on 
the  wall? 

The  weight  of  water  reaching  the  wall  per  second  is  rep- 
resented by  a  cylinder  of  water  2  in.  diameter  and  200  ft. 

1       I      22 

long=—  X  —  X—X  200X62.5  =  273  Ibs.      The     change     of 

velocity  is  equal  and  opposite  to  the  velocity  of  the  stream, 
or  200  ft  /sec.     Therefore  reaction  of  wall 

=  273X200 
32.2X1 

Example  2.  —  120  cu.ft.  of  water  leave  the  rim  of  the 
wheel  of  a  centrifugal  pump  every  minute.  The  component 
velocity  of  the  water  in  the  direction  of  motion  of  the 
rim  is  20  ft  /sec.,  and  the  velocity  of  the  rim  is  25  ft  /sec. 
If  the  water  enters  at  the  center  with  no  velocity  in  .the 


196  ELEMENTARY  PRACTICAL  MECHANICS 

direction  of  the  rim,  what  H.P.  will  be  required  to  drive 
the  pump,  neglecting  friction? 

120  cu.  ft.  per  minute  =  2  cu.  ft.  each  second  =  2X62.5  = 
125  Ibs.  of  water  each  second.  To  impart  to  this  a  velocity 

125X20 

of  20  ft /sec  requires  a  force  of =77.6  Ibs. 

32. 2  X 1 

If  the  velocity  of  the  rim  is  25  ft/sec,  the  work  required 
is  77.6X25  =  1940  ft.-lbs.  per  second.  The  H.P.  required 
is  therefore 

1940X60 


33,000 


-  =  3.53  H.P. 


PROBLEMS 

1.  A  truck  weighing  12  tons  moving  with  a  velocity  of 
5  ft./ sec  is  stopped  by  buffers  in  ^  second.     What  is  average 
force  of  the  blow? 

2.  A  ship  of  10,000  tons  moving  at  5  miles  per  hour  is 
stopped  in  one  minute.     Average  force  required? 

3.  A  gun  delivers  100  bullets  per  minute  each  weighing 
one  ounce,  with  a  horizontal  velocity  of  1610  ft/sec.     What 
is  the  average  force  exerted  upon  the  gun? 

4.  A  1J  in.  stream  from  a  fire-engine  is  thrown  horizon- 
tally against  a  wall  with  a  velocity  of  150  ft/sec.     What 
force  is  exerted  against  the  wall? 

5.  A  centrifugal  pump  discharges  80  cu.  ft.  of  water  per 
minute  at  a  velocity  tangent  to  the  rim  of  the  wheel  of 
25  ft/sec.     If  the  velocity  of  the  rim  is  30  ft/sec,  what 
is  the  H.P.  required? 

104.  Principle    of    the    Conservation     of     Momentum. 

Since  a  force  is  measured  by  the  momentum.it  will  gen- 
erate in  unit  time,  it  follows  that  equal  forces,  acting  for  the 
same  time,  will  generate  equal  momenta.  The  momentum 


ELEMENTARY  PRACTICAL  MECHANICS         .    197 

imparted  to  a  gun  when  fired,  if  the  gun  can  be  considered 
as  free  from  other  bodies,  will  equal  the  momentum  im- 
parted in  the  opposite  direction  to  the  bullet,  although 
because  of  its  greater  mass  the  velocity  of  the  gun  will  be 
very  small  in  comparison;  and  in  general,  the  same  is 
true  for  any  system  of  two  bodies  acted  upon  simulta- 
neously by  equal  and  opposite  forces.  When,  for  example, 
a  body  A  strikes  a'nother  body  B,  the  mutual  pressures 
between  the  bodies  are  equal  and  opposite  during  the  time 
of  impact.  If  A  loses  momentum,  B  must  gain  an  equal 
amount  provided  the  action  of  outside  bodies  upon  both, 
can  be  neglected.  The  total  momentum  of  the  system 
(i.e.,  of  A  and  B  taken  together)  therefore  remains  the 
same  as  it  was  before  impact,  although  the  velocities  of 
both  A  and  B  may  be  changed  in  either  amount  or  direc- 
tion or  in  both.  This  fact  that  the  total  momentum  of  a 
system  remains  constant  unless  a  force  external  to  the  system 
is  applied  is  known  as  The  Law  of  Conservation  of  Momen- 
tum. 

It  must  not  be  inferred  from  what  has  been  said,  how- 
ever, that  the  total  mechanical  energy  of  the  system  remains 
constant  also.  During  impact,  work  is  done  against  inter- 
nal forces  (cohesion,  internal  friction,  etc.),  and  some 
mechanical  energy  is  thus  transformed  into  heat  energy. 
The  mechanical  energy  is  therefore  always  less  after  impact. 


CHAPTER  XI 


1 


FRICTION.    THE  GENERAL  LAWS  OF  MACHINES 

105.  Friction. — The  resistance  offered  to  the  sliding  of 
one  body  on  another  is  called  the  force  of  friction  between 
the  two  bodies. 

Thus,  suppose  a  cast-iron  block  weighing  W  pounds 
rests  upon  the  horizontal  surface  AB,  also  of  cast  iron, 
Fig.  145.  The  pressure  between  the  surfaces  in  contact 

will  be  W  pounds.  A  force  Si 
will  be  required  to  cause  the 
body  W  to  start  from  rest;  a 
smaller  force,  $2,  will  keep  the 
body  moving  uniformly  when  once 
started.  Si  therefore  measures 
the  force  of  starting  friction  or 
static  friction  as  it  is  usually 
called,  and  82  measures  the  force 
of  friction  of  motion  or  sliding 
friction.  If  the  pressure  W  between  the  surfaces  be  increased 
by  adding  weights  to  the  block,  a  larger  force,  Si,  will  be 
required  to  start  the  block  than  was  needed  before,  and  also 
a  larger  steady  pull,  S2,  to  keep  it  sliding  steadily.  In  air 
cases,  however,  Si  will  be  greater  than  82,  or  static  friction 
is  greater  than  sliding  friction. 

If  a  series  of  determinations  of  both  sliding  and  static 
friction  be  made  for  different  pressures  between  the  sur- 
faces in  contact,  the  results  may  best  be  shown  graphically, 
as  in  Fig.  146.  The  resulting  curves  will  be  straight  lines 
passing  through  the  origin.  They  show  therefore  that  for 

198 


FIG.  145. 


ELEMENTARY  PRACTICAL  MECHANICS 


199 


both  static  and  sliding  friction,  the  force  of  friction  is  pro- 
portional to  the  pressure  perpen- 
dicular to  the  surfaces  in  contact , 
and  therefore  that: 


FRICTION 


PRESSURE 


=  a  constant. 


f 


Pressure  in  Lbs. 

FIG.  146. 


If  we  change  the  area  of  the  I 
surfaces  in  contact,  keeping  the 
total  pressure  between  them  un- 
changed, as  for  example,  if  we 
place  W  upon  its  narrow  side, 
we  will  obtain  practically  the 
same  values  for  the  force  of  friction  provided  this  side  has 
been  dressed  in  the  same  manner  as  the  other.  The  force 
of  friction  is  therefore  practically  independent  of  the  area 
of  the  surfaces  in  contact,  where  cutting  does  not  occur. 

106.  Coefficient  of  Friction. — The  fraction,   - —  is 

called  the  Coefficient  of  friction. 


Pressure 


Static  friction 


Pressure  normal  to  surface 


=  Coefficient  of  friction  of  rest. 


Sliding  friction 
Pressure  normal  to  surfaces 


=  Coefficient  of  friction  of  motion. 


The  coefficient  of  sliding  friction,  usually  called  simply  the 
coefficient  of  friction,  is  the  one  commonly  required  in  prac- 
tice, and  is  represented  by  the  symbol  //. 

If  instead  of  cast-iron  surfaces  we  use  brass  on  cast  iron, 
cast  iron  on  leather,  oak  on  oak,  fibers  parallel,  oak  on  oak, 
across  the  grain,  etc.,  -we  will  still  obtain  curves  which  are 
straight  lines  passing  through  the  origin,  but  each  curve 
will  have  a  different  slope.  In  other  words,  the  fraction 


200  ELEMENTARY  PRACTICAL  MECHANICS 

.p: will  be  a  constant  for  each  case,  but  will  have  dif- 

Pressure 

ferent  values  for  each  case.  The  coefficient  of  friction 
therefore  expresses  the  effects  of  material,  surface,  etc., 
upon  friction.  The  student  will  find  it  interesting  here  to 
compare  the  values  of  the  coefficients  of  friction  for  differ- 
ent materials  given  in  the  larger  reference  books  on  me- 
chanics. 

-n  j  r   -^  Friction  ,.    L   .  . 

.brom  our  definition  IJL  =  ^ ,  we  see  that,  in  general, 

Pressure 

FORCE  TO  SLIDE  ONE  BODY  ON  ANOTHER  =  /JL  X  NORMAL  PRES- 
SURE BETWEEN  THE  SURFACES. 

107.  Determination    of    Coefficient    of    Friction. — The 

coefficient  of  friction  between  two  surfaces  mav  be  obtained 


FIG.  147. — Laboratory  Inclined  Plane. 

as  in  Fig.  145.  A  more  convenient  form  of  apparatus  is 
shown  in  Fig.  147.  CD  is  an  inclined  plane  which  can  be 
adjusted  to  any  desired  angle  of  slope.  Upon  this  are 
placed  the  surfaces  to  be  tested,  A  and  B,  A  being  free  to 
move.  The  weight,  W,  of  A  may  be  resolved  into  the  two 


ELEMENTARY  PRACTICAL  MECHANICS 


201 


forces  P,  tending  to  cause  A  to  slide  on  J3,  and  R,  perpen- 
dicular to  the  line  of  contact  between  A  and  J3,  producing 
pressure  between  them. 

Rough  adjustment  of  CD  is  made  by  moving  the  tri- 
angular block  T  forward  or  back.  Fine  adjustment  is  then 
made  by  tipping  the  block  by  turning  the  screw  S  until,  on 
tapping  A  to  start  it,  it  slides  uniformly  on  B. 

P    CE 

Thus  friction  =  P  and  //=—==——==  tangent  of   angle   of 

plane  CDE. 

(Student  should  prove  geometrically  that  u  =  tangent  of 
angle  of  plane.) 

108.  Friction  of  Machinery.  —  If  in 
place  of  the  apparatus  shown  in  Fig.  145, 
a  simple  machine,  as  the  wheel  and  axle  of 
Fig.  148,  be  used,  the  value  of  E  when 
W  is  raised  steadily  must  be  enough  to 
balance  W  and  also  overcome  friction  of  the 
apparatus:  When  W  is  lowered  steadily, 
E=  force  to  balance  W  minus  friction. 
Calling  force  to  balance  W  on  a  friction- 
less  machine  P,  we  thus  have  :  FIG.  148. 


E,  W  rising 

E,  W  descending  =  P  —  Fr. 

From  these  we  see  that  friction,  Fr 

Difference  between  the  two  values  of  E 


Plotting  results  obtained  with  different  loads,  W,  will  give 
the  straight  line  of  Fig.  149.  This  line  gives  an  intercept 
K,  on  the  Y  axis,  equal  to  the  friction  of  the  machine  unloaded, 
and  therefore  its  equation  is,  Friction  =  friction  of  unloaded 
machine  +  increase  in  friction  due  to  load  (equal  to  slope, 
c,XW).  Or, 

(1)  Fr=K+cW; 


202 


ELEMENTARY  PRACTICAL  MECHANICS 


Or,  substituting  values  from  our  curve, 
(2)  Fr=l  +  .05Xload. 

Equation  1  is  a  perfectly  general  expression  for  the 
friction  in  machines  whether  they  are  large  or  small,  sim- 
ple or  complex.  It  should  be  noted,  however,  that  the 
value  of  friction  is  here  measured  at  the  circumference  of  the 
wheel.  Of  course  the  actual  resistance  occurs  in  the  bear- 
ings and  is  probably  distributed,  occurring  at  a  number  of 

points.  It  is  entirely 
feasible,  to  measure  it  in 
terms  of  force  required 
to  turn  the  apparatus 
with  a  moment  arm  equal 
to  the  radius  of  the  wheel. 
It  might  have  been  de- 
termined in  terms  of 
force  applied  to  the  cir- 
cumference of  the  axle, 
or  at  any  other  conveni- 
ent point,  if  so  desired.  Thus  in  machinery,  a  part  of  the 
driving  force,  wherever  it  is  applied,  is  required  to  over- 
come the  friction  distributed  through  the  machine,  and  the 
amount  of  force  necessary  for  this  purpose  depends  first 
upon  the  friction  of  the  unloaded  machine  and  secondly 
upon  the  increase  in  friction  produced  when  load  is  put  upon 
the  parts.  In  general,  therefore,  friction  is  a  disadvantage 
in  machines,  diminishing  their  efficiency  as  we  shall  see 
later.  In  a  few  special  machines,  as  for  instance  a  screw- 
jack,  friction  is  necessary  for  their  proper  operation. 

109.  So-called  Laws  of  Friction. — In  the  preceding  arti- 
cles, to  make  the  points  under  discussion  more  clear,  we 
have  spoken  of  the  friction  of  different  surfaces,  coeffi- 
cients of  friction,  etc.,  as  if  they  were  fixed,  and  accurately 


Friction  in  Lbs. 

K>  M»  C 

x^ 

.x^ 

x^ 

x" 

X 

x' 

x^ 

x^ 

X 

X 

JL 

)              20              40              60              80    ,          10 
Added  Load  In  Lbs. 

FIG.  149. 

ELEMENTARY  PRACTICAL  MECHANICS 


203 


known  values  like  the  constants  used  in  all  science  work. 
As  a  matter  of  fact,  the  factors  controlling  friction  in  any 
actual  instance  are  so  numerous,  and  so  dependent  upon 
variable  conditions  that  only  the  most  general  principles 
may  be  stated  positively.  The  following  summary  is 
taken  from  Perry's  "Apulied  Mechanics." 


LAWS  OF  FRICTION 


FRICTION  BETWEEN  SOLIDS. 


FLUID  FRICTION. 


1.  The  force  of  friction  does  not 

much  depend  on  velocity  but 
is  greatest  at  slow  speeds. 

2.  The  force  of  friction  is  propor- 

tional to  the  total  pressure  be- 
tween the  two  surfaces. 

3.  The   force   of  friction   is   inde- 

pendent of  the  areas  of  the 
rubbing  surfaces. 

4.  The  force  of  friction  depends 

very  much  on  the  nature  of 
the  rubbing  surfaces,  their 
roughness,  etc. 


4. 


The  force  of  friction  very  much 
depends  on  the  velocity,  and 
is  indefinitely  small  when  the 
speed  is  very  slow. 

The  force  of  friction  does  not 
depend  on  the  pressure. 

The  force  of  friction  is  propor- 
tional to  the  area  of  the  wetted 
surfaces. 

The  force  of  friction  at  moderate 
speeds  does  not  much  depend 
on  the  nature  of  the  wetted 
surfaces. 


From  Perry's  "  Applied  Mechanics,"  page  80. 

110.  Friction   of  Lubricated  Surfaces. — The  very  great 

differences    between     the      6 
friction  of  dry  bodies  and    ^ 
of  fluids,  as  stated  in  the    |4 
preceding  article,  should  be    5s 
noted.       Even    with    dry    | 

surfaces  the  amount  of  air   * 

• 

film  between  the  surfaces    I1 
may    greatly  modify    the 
friction.     Lubricated  sur- 
faces,   as    ordinary    bear- 


1000  6000 

Load  in  Pounds 


FIG.  150.— Starting  Torque. 


204 


ELEMENTARY  PRACTICAL   MECHANICS 


ings,  etc.,  would  naturally  obey  neither  the  laws  for  dry 
surfaces  nor  those  for  fluids,  but  would  be  intermediate 
between  the  two,  varying  from  the  case  of  flooded  bearings 
which  follow  quite  nearly  the  laws  for  fluids,  to  very  modi- 
fied and  uncertain  conditions  as  the  degree  of  lubrication, 
kind  of  lubricant,  temperature,  pressure  between  surfaces, 
velocity,  etc.,  change.  No  very  definite  statements  can 
therefore  be  made  for  the  friction  of  lubricated  surfaces. 

In  general,  the  character  of  the  lubricant  should  be  such 
that  it  will  be  constantly  dragged  into  the  rubbing  sur- 
faces as  needed.  Where 
the  pressure  between  the 
surfaces  is  great,  light, 
thin  oils  would  be  forced 
out,  hence  for  such  cases 
heavy  oils  or  grease  are 
preferable.  Under  light 
pressures  these  would  be 
too  viscous. 

Load  m  Pounds 


Horse  Power  Consumed  in  Frictio 
O  I*.  .  be  „  to  b>  c 

x** 

Br 

onze 

/ 

,x* 

X*"0 

Cast 

Iron 

9' 

^ 

o 

/\ 

X^' 

4 

R 

jller 

rf 

T— 

•0- 

2000           4:000           6000           8000 

FIG.  151. — Horse  Power  consumed  in 
Friction.  Speed  130  revolutions 
per  minute. 


The  properties  of  oils 
vary  greatly  with  temper- 
ature also,  and  the  selec- 
tion of  a  lubricant  must 

depend  considerably  upon  the  temperature  to  which  it  will 
be  subjected.  For  high  temperatures  such  as  result  when 
high-pressure  steam  is  used,  mineral  oils  which  do  not 
decompose  with  heat  must  be  used. 

The  horsepower  expended  in  driving  a  shaft  =  force  of 
friction  at  the  given  speed  XxX  diam.  X  r.p.m.  -*-  33,000 ;  or, 
what  amounts  to  the  same  thing, 

H.P.  =Coef.  friction  at  the  given  speed  X  pressure  of 

shaft  in  bearings  X  xX  diam.  X  r.p.m -f-  33, 000. 

Poor  alignment  greatly  increases  the  pressure  of  shaft  in 


ELEMENTARY  PRACTICAL  MECHANICS 


205 


[orse  Power  Consumed  in  Friction 

j-  ta  jo  '  ~  p-  j 
o  oo  o.o  < 

\ 

/ 

7 

Jrons 

e  / 

/ 

>r 

f^ 
t 

^ 

* 

/ 

,^'' 

LC* 

tiro 

i 

. 

I 

rf 

~? 

—  R< 

•"6 

•^ 

,  —  °- 

,-0— 

^ 

--  o—  ' 

2000        ,  4000     .6000           8000 

its  bearings  and  therefore  the  friction.  Any  computation 
based  upon  this  equation,  however,  is  necessarily  only 
approximate  owing  to  lack  of  knowledge  of  the  friction  at 
a  particular  speed.  Figs.  151  and  152  show  the  results  of 
tests  of  horsepower  required  to  keep  a  shaft  turning  at 
constant  speed  in  bronze,  cast-iron  and  roller  bearings  at 
130  r.p.m.  and  460  r.p.m. 
with  the  same  loads.  At 
a  speed  approximately  3.5 
times  as  fast,  the  horse- 
power for  cast  iron  is  only 
2.5  and  for  roller  bearings 
2.9  times  as  much;  or  in 
other  words,  in  these  cases 
the  friction  is  less  at  a 
high  speed.  In  the  case 
of  the  bronze  bearing,  on 
the  contrary,  the  horse- 
power given  by  the  test 

was  5  times  as  much,  showing  an  apparent  increase  of  friction 
with  the  speed.  No  divisions  of  mechanics  furnish  a  better 
medium  for  instructive  experiments  than  those  dealing  with 
the  relation  of  friction  to  material,  speed,  lubrication,  etc., 
and  with  the  power  required  to  drive  a  shaft  under  different 
conditions  of  lubrication,  alignment,  etc.  The  student 
should,  therefore,  be  given  ample  opportunity  for  direct, 
practical  tests  upon  a  shaft  driven  by  a  variable  speed  motor. 
Friction  brakes  and  transmission  dynamometers  for  use 
in  such  tests  are  described  in  Art.  116. 

111.  Friction  of  Belts. — Suppose  a  pulley  P\  is  driving 
a  second  pulley  P^  by  means  of  a  belt,  the  direction  of 
rotation  being  as  shown  by  the  arrows,  Fig.  153;  both 
sides  of  the  belt,  t  and  T,  must  be  under  tension  in  order 
to  give  pressure  on  the  pulleys,  and  therefore  friction  which 


Load  in  Pounds 

FIG.  152. — Horse  Power  consumed 
in  Friction.  Speed  460  revolu- 
tions per  minute. 


206  ELEMENTARY  PRACTICAL  MECHANICS 

shall  keep  the  belt  from  slipping.  But  tension  T  must 
be  greater  than  the  tension  t,  otherwise  P2  will  not  turn. 
It  is  this  difference  in  tension  of  the  two  sides  of  the  belt 
therefore  which  measures  the  force  involved  in  the  trans- 
mission of  power.  Difference  in  tension  X  speed  of  belt  in 
ft.  per  min.  =  work  done  per  minute.  And 

difference  in  tension  X  speed  in  ft.  per  min.  _ 
33,000 

transmitted  from  PI  to  P^. 

The  friction  of  a  belt  on  the  pulley  may  be  increased  by 
increasing  the  tension — and  therefore  the  pressure  between 

the  belt  and  pulley — or  by 
increasing  the  arc  of  con- 
tact between  the  two.  The 
usual  practice  is  to  drive 
with  the  upper  side  of  the 

T  j ^ — -^          belt  slack,  so  that  any  sag- 

FIG.  153.  ging  due  to  the  weight  of 

the  belt  may  increase  the 

arc  of  contact.  "Idlers"  or  tightening  pulleys  are  often 
placed  on  the  upper  or  slack  side  of  the  belt  to  increase 
both  the  tension  and  the  arc  of  contact. 


PROBLEMS 

1.  A  block  weighing  200  Ibs.  rests  on  a  horizontal  sur- 
face.    It  requires  90  Ibs.  pull  to  slide  it  along  with  uniform 
speed.     Find  the  coefficient  of  friction  and  the  work  done 
in  moving  it  20  ft. 

2.  A  stone  is  projected  along  smooth  ice  with  a  velocity 
of  15  ft.  per  second  and  comes  to  rest  after  traveling  75 
yards.     Find  the  coefficient  of  friction  of  stone  on  ice. 


ELEMENTARY  PRACTICAL  MECHANICS  207 

3.  If  it  requires  a  force  of  65  Ibs.  to  move  a  body  weigh- 
ing 100  Ibs.  uniformly  up  a  plane,  inclined  30°,  find  coeffi- 
cient of  friction  for  the  surfaces. 

4.  Find  the  work  done  in  sliding  a  wooden  body  weighing 
300  Ibs.  up  an  inclined  plane  5  ft.  high  and  20  ft.  long,  the 
coefficient  of  friction  being  0.4. 

5.  A  block  of  stone  is  hauled  along  the  ground  by  a  pull 
of  310  Ibs.  at  an  angle  of  10°  above  the  horizontal.     If 
coefficient  of  friction  is  0.58,  what  is  the  weight  of  the  stone? 

6.  A  wagon  weighing  2200  Ibs.  is  pulled  up  a  hill  rising 
2  ft.  in  20  ft.  of  road.     If  the  resistance  is  12  Ibs.  per  hun- 
dred weight,  what  pull  parallel  to  the  load  must  be  exerted? 

7.  What  is  the  least  pull  which  will  move  a  box  weighing 
200  Ibs.  along  a  concrete  floor  if  the  coefficient  of  friction 
is  0.50? 

NOTE. — Least  pull  acts  at  an  angle  above  horizontal  whose 
tangent  =  coefficient  of  friction. 

8.  A  ladder  30  ft.  long  weighing  100  Ibs.  rests  against  a 
wall.     Center  of  gravity  of  ladder  is  J  the  distance  from 
the  foot.     Coefficient  of  friction  vfiih  ground  is  0.6  and 
with  wall  above  is  0.3.     Ladder  makes  angle  of  45°  with 
the  wall.     How  far  up  the  ladder  can  a  man  weighing  200 
Ibs.  go  before  the  ladder  begins  to  slip? 

9.  A  cast-iron  fly-wheel  has  a  rim  6  ft.  mean  diameter, 
7  ins.  wide,  and  4  ins.  thick.     It  is  keyed  to  a  shaft  4  ins. 
diameter,  weighing  500   Ibs.     If  the  fly-wheel  is  running 
148  revolutions  per  minute  and  the  coefficient  of  friction 
between   shaft   and   bearing   is   0.05,    approximately   how 
many  revolutions  will  the  fly-wheel  make  in  coming  to  rest? 

10.  What  horsepower  will  be  required  to  drive  the  loaded 
shaft  of  Problem  9,  at  the  uniform  speed  of  240  revolutions 
per  minute? 

11.  If  the  coefficient  of  friction  of  carbon  on  copper  is 
0.24  and  if  the  average  pressure  of  the  10  sq.in.  of  carbon 
upon  a  14  in.  diameter  commutator  of  a  motor  is  3  Ibs. 
per  sq.  in.,  what  will  be  the  horsepower  used  in  overcoming 
friction  at  a  speed  of  600  revolutions  per  minute? 


208  ELEMENTARY  PRACTICAL  MECHANICS 

12.  A  N.  Y.  C.  locomotive  weighing  200  tons  and  exert- 
ing a  tractive  force  of  20,000  Ibs.  hauls  a  trailing  load  of 
400  tons.     If  friction  is  6^  Ibs.  per  ton  how  long  will  the 
locomotive  require  to  get  up  a  speed  of  55  miles  per  hour 
from  rest:    (a)  On  a  level?    (6)  On  a  2%  grade? 

13.  What  is  the  maximum  trailing  load  that  the  loco- 
motive of  problem  12  can  keep  moving  at  a  constant  speed 
of  45  miles  per  hour  up  a  1%  grade? 

112.  Law  of  Conservation  of  Energy. — This  law,  briefly 
stated,  is: 

1.  Energy    cannot    be    created    or    destroyed.     The 
total  sum  of  energy  in  the  universe  remains  constant. 

2.  Energy   may  be  transformed  from   one  form  to 
another;  in  such  transformation  no  energy  is  ever  lost. 

These  propositions  have  been  established  by  observa- 
tions and  experiments  extending  over  many  years,  and 
conducted  by  men  working  in  all  branches  of  natural 
science,  especially  in  physics  and  chemistry.  They  are  now 
universally  accepted  and  form  the  fundamental  principle 
for  all  theories  not  only  in  engineering,  and  physics  and 
chemistry,  but  also  in  physiology,  botany,  astronomy,  and 
in  fact  every  natural  science.  The  principle  of  the  Con- 
servation of  Energy  thus  becomes  both  an  aid  to  the  under- 
standing of  many  natural  processes  and  a  test  by  which 
science  determines  the  reasonableness  of  each  new  theory 
or  hypothesis. 

The  final,  definite  statement  of  the  law  of  the  conserva- 
tion of  energy  dates  from  the  middle  of  the  nineteenth 
century,  when  it  was  experimentally  proved  that  heat  is 
a  "  form  of  energy,"  that  is  to  say,  A  CERTAIN  AMOUNT  OF 

HEAT  MAY  BE  MADE  TO  PERFORM  A  DEFINITE  AMOUNT  OF 
WORK,  AND  CONVERSELY  A  CERTAIN  AMOUNT  OF  MECHANI- 
CAL WORK,  IF  TRANSFORMED  INTO  HEAT,  WILL  FURNISH  A 


ELEMENTARY  PRACTICAL  MECHANICS  209 

DEFINITE  AMOUNT  OF  HEAT.     This  fact  may,  for  our  pur- 
pose, be  stated  thus: 

THE  ENERGY  REQUIRED  TO  HEAT  1  LB.  OF  WATER  ONE 
DEGREE  FAHRENHEIT,  IS  THE  SAME  AS  THAT  REQUIRED  .TO 
PERFORM  778  FT.-LBS.  OF  WORK. 

Of  course  the  equivalence  of  various  "  forms  of  energy  " 
may  be  stated  in  various  other  ways,  which  will  include 
such  apparently  different  ideas  as  electrical  and  chemical 
relations  and  the  phenomena  of  sound,  light,  and  animal 
and  vegetable  processes.  Thus  take  1000  ft.-lbs.  of  mechan- 
ical energy  and  by  proper  mechanism,  either  natural  or  of 
human  devising,  transform  it  into  other  forms  of  energy; 
we  find  that  always  a  certain  perfectly  definite  amount  of 
chemical  or  electrical  or  other  energy  is  obtained  and  that 
by  no  device  can  we  ever  produce  more  energy  than  the 
1000  ft.-lbs.  we  started  with.  It  is  perhaps  unnecessary 
to  add  that  the  complete  tracing  of  these  transformations 
and  the  measurement  of  the  amounts  at  all  stages  of  the 
processes  is  not  humanly  possible,  since  the  processes  are 
as  intricate  and  manifold  as  nature  itself.  All  our  ex- 
periments, however,  point  undeniably  to  the  truth  of 
the  great  fact  that  energy  may  be  transformed  but  never 
destroyed. 

It  must  not  be  inferred  that  the  law  of  conservation  of 
energy  means  that  the  supply  of  energy  available  for  our 
use  is  constant.  In  all  machinery  by  which  energy  is  util- 
ized, there  is  a  constant  loss  of  useful  energy  through  the 
transformation  into  heat  energy  by  friction  at  bearings, 
etc.,  and  the  dissipation  of  this  heat.  Thus,  as  an  illus- 
tration, suppose  10,000  units  of  heat  energy  are  liberated 
by  the  combustion  of  1  Ib.  of  coal  in  the  boiler  furnace. 
Of  this  about  7500  units  are  given  to  the  steam,  the  other 
2500  units  are  dissipated  through  the  ash,  chimney  gases, 
radiation,  etc.  Of  the  7500  units  carried  into  the  cylinder 


210  ELEMENTARY  PRACTICAL  MECHANICS 

about  800  units  are  transformed  into  mechanical  energy 
of  the  moving  piston,  6700  units  being  lost  through  con- 
densation, radiation,  etc.,  in  the  cylinder,  or  carried  on 
into  the  condenser.  If  the  engine  is  driving  a  dynamo, 
about  700  of  the  800  units  may  reappear  in  the  energy  of 
the  electric  current,  the  other  100  disappearing  as  heat 
through  friction,  electrical  losses,  etc.  If  the  dynamo  is 
supplying  arc  or  incandescent  lights,  the  700  units  are 
given  to  heat  the  connecting  wires,  lamp  filaments,  etc., 
and  thus  all  our  original  supply  of  energy  has  been  dissi- 
pated. Or,  if  the  dynamo  drives  a  motor,  about  650  units 
may  be  obtained  in  mechanical  energy  supplied  by  the 
motor,  50  being  lost  in  the  transformation  through  friction, 
electrical  losses,  etc.  Of  the  total  10,000  units  with  which 
we  started,  therefore,  approximately  94.5%  have  been 
dissipated,  6.5%  are  still  available  for  doing  work. 

Many  similar  examples  might  be  given.  In  all  we  find 
that  each  step  in  the  process  of  utilizing  energy  is  accom- 
panied by  unavoidable  loss. 

113.  Machines. — For  students  of  Mechanics  the  imme- 
diate and  most  useful  application  of  the  Principle  of  the 
Conservation  of  Energy  is  in  the  study  of  machines.  In 
the  most  general  terms  we  may  say  that,  A  machine  is  a 
device  by  which  energy  received  from  some  source  outside 
itself  is  delivered  at  some  other  point,  after  certain  "  losses  " 
and  transformations,  there  to  perform  some  special  work. 

Thus  with  a  derrick,  the  energy  supplied  to  the  hoist 
serves  to  lift  weights  against  gravity.  An  engine  is  a 
machine  in  which  the  heat  energy  of  the  steam  is  trans- 
formed, with  certain  "  losses/7  into  the  kinetic  energy  of 
rotation  of  the  fly  wheel. 

It  will  be  seen  that  the  law  of  the  conservation  of  energy 
furnishes  us  with  certain  necessary  specifications  for  the 
complete  study  of  any  machine.  From  such  study  we  may 


ELEMENTARY  PRACTICAL  MECHANICS  211 

arrive  at  a  determination  of  a  machine's  efficiency,  that  is, 
of  the  per  cent,  of  the  energy  supplied  that  is  actually  em- 
ployed in  the  useful  work  for  which  the  machine  is  designed. 

At  the  driving  end  a  more  or  less  constant  stream  of 
energy  is  supplied  from  the  driving  belt,  shafting,  moving 
water,  steam,  Nele€tric  motor,  etc.  None  of  this  energy 
received,  or  "  input,"  can  be  destroyed;  that  which  does 
not  pass  through  and  reappear  in  the  output  of  the  machine, 
has  been  required  for  internal  losses  which  have  dissipated 
it  as  non-recovered  heat.  Each  moving  part  of  the  ma- 
chine plays  its  part  in  this  transmission  of  energy.  Each 
has  its  own  store  of  energy  which  is  constantly  augmented 
by  the  supply  received  from  the  preceding  part  of  the 
mechanism,  and  diminished  by  the  energy  handed  on  to  the 
next  following  part  with  which  it  is  joined.  Starting  at 
the  driving  end,  a  debit  and  credit  sheet  might  be  kept 
for  each  succeeding  unit  in  the  mechanism;  and  in  this 
every  part  would  be  found  unable  to  repay  the  energy  re- 
ceived in  full  because  of  the  inevitable  operating  losses. 

The  designer  may  reduce  the  waste  of  energy  at  any  par- 
ticular point  by  devices  for  lessening  friction,  and  by  prop- 
erly proportioned  bearings,  etc.,  which  may  dimmish  the 
distance  through  which  friction  must  be  overcome.  He 
may  modify  a  fluctuating  energy  supply  into  a  compara- 
tively steady  'output  by  fly-wheels  attached  to  certain 
parts,  which  on  account  of  their  great  capacity  for  storing 
kinetic  energy,  absorb  a  great  supply  without  great  in- 
crease in  speed,  and  then  when  the  supply  temporarily 
diminishes  or  great  demand  is  suddenly  made  on  the  part, 
can  supply  a  large  quantity  of  energy  without  greatly 
^slowing  up;  or  he  may  attach  a  governor  to  a  steam  engine 
or  a  water  wheel  to  automatically  adjust  the  supply  of 
energy  to  meet  the  varying  demands  for  energy  made  by 
the  machinery  in  operation.  But  for  any  form  of  machine, 


212  ELEMENTARY  PRACTICAL  MECHANICS 

whether  simply  mechanical  or  involving  the  use  of  steam, 
water,  electrical  or  other  motive  power,  he  must  work  in 
accordance  with  the  principle  of  the  conservation  of  energy, 
for  he  cannot  supplant  it.  He  must  allow  for  the  inevi- 
table losses,  realizing  that  not  all  of  the  energy  supplied  can 
be  delivered  for  performing  useful  work. 

Thus  machine  design  becomes  primarily  a  study  of 
energy  relations  and  energy  losses.  The  "  explanation  " 
of  the  operation  of  any  machine  involves  a  recognition  of 
the  energy  supply,  the  mechanism  and  processes  of  energy 
transformations,  sources  of  energy  loss,  and  the  fact  that 
the  energy  finally  delivered  for  the  purpose  desired  is  only 
a  fraction  of  the  energy  received. 

114.  Important  Terms  Applied  to  Machines. — The  follow- 
ing definitions  are  important  in  connection  with  machines: 

1.  INPUT. — The  energy  supplied  TO   the  machine  is 
known  as  the  INPUT. 

2.  OUTPUT. — The  energy  delivered  BY  the  machine  for 
the  purposes  for  which  it  was  designed  is  known  as  the 
OUTPUT. 

3.  EFFICIENCY. — The  efficiency  of  a  machine  equals 
the  ratio  of  output  to  input.     It  is  usually  expressed  as 
a  per  cent,  i.e.,  per  cent  of  input  which  is  delivered  by 
the  machine.     Thus: 

.  output 

Efficiency  =- , 

input 

and 

Output  =  input  X  efficiency. 

4.  VELOCITY  RATIO. — The  distance  through  which  the 
driving  force  acts  divided  by  the  distance  through  which 
the  force  exerted  by  the  machine  acts  in  the  same  time,  is 
known  as  the  VELOCITY  RATIO. 


ELEMENTARY  PRACTICAL  MECHANICS  213 

Thus  referring  to  the  simple  wheel  and  axle  shown  in 
Fig.  148,  if  circumference  of  wheel  A  is  48  ins.  and  of  axle 
B  is  15  ins.,  and  the  apparatus  is  used  for  hoisting  loads 
applied  at  TF,  for  each  revolution  the  force  applied  E 
moves  48  ins.,  the  force  exerted  by  machine,  or  W,  moves 

15  ins.;   the  velocity  ratio  is  thus  —  =  3.2. 

lo 

The  velocity  ratio  is  thus  a  question  of  the  design  of  the 
machine. 

5.  MECHANICAL  ADVANTAGE. — The  ratio  of  force 
exerted  by  a  machine  to  driving  force  is  known  as  the 
MECHANICAL  ADVANTAGE  of  the  machine. 

Thus,  referring  again  to  Fig.  148, 

W 

Mechanical  advantage  =  —. 

Ei 

It  is  obvious  that  if  the  machine  could  be  frictionless 
output  would  equal  input,  and  work  done  on  W  would  equal 
work  done  by  E,  or  TF  X  circumference  of  B  =  E  X  circum- 

W    48 

ference  of  A.     Therefore,  — -  = —  =  3.2,  or  mechanical  ad- 

E      15 

vantage  would  equal  the  velocity  ratio.  In  any  actual 
machine,  however,  because  of  friction,  the  driving  force 
must  be  increased  from  that  required  for  a  frictionless 
machine,  hence  the  actual  mechanical  advantage  will  always 
be  less  than  the  velocity  ratio.  It  will  also  be  variable  with 
the  conditions  of  lubrication,  bearing  surfaces,  etc.,  and 
must  therefore  be  determined  by  actual  experiment  at 
the  time  the  machine  is  used. 

115.  General  Law  of  Machines. — From  the  preceding 
discussion  it  will  be  seen  that  the  general  law  for  any  type 
of  machine  will  be  of  the  form: 


214  ELEMENTARY  PRACTICAL  MECHANICS 

Input  =  output  +  energy  expended  against  friction. 
Or, 


Where  E  =  driving  force,  L  =  force  applied  by  machine 
—  commonly  called  Load  —  Fr  =  force  to  overcome  friction, 
and  di,  d2,  d3  are  the  distances  through  which  driving 
force,  load,  and  friction  act,  respectively. 

Assuming  no  friction,  as  is  sometimes  convenient,  this 
equation  becomes, 

Input  =  output      or      Exdi=Lxd2. 

In  the  study  of  any  actual  machine  where  friction  is 
always  present  and  therefore  some  of  the  input  is  not 
recoverable,  a  convenient  expression  of  the  law  is: 

Input  X  efficiency  =  output, 
or 

E  X  di  X  efficiency  =  L  X  d2. 

116a.  Dynamometers.  —  Instruments  for  measuring  power 
are  commonly  called  dynamometers.  These  are,  generally 
speaking,  of  two  classes:  (1)  absorption  dynamometers  which 
absorb  the  power  received,  transforming  it  into  heat;  and 
(2)  transmission  dynamometers  which  absorb  only  power 
enough  to  operate  their  mechanism  and  transmit  the 
remainder. 

Dynamometers  of  the  first  type  are  used  where  the  out- 
put of  a  source  of  power  is  to  be  measured,  as  for  example, 
the  power  delivered  at  the  belt  pulley  of  a  motor,  engine, 
etc.  This  is  termed  the  brake  horsepower  of  the  motor  or 
engine. 

Those  of  the  second  type  are  inserted  between  the  source 
of  power  and  the  machine  being  driven  to  determine  the 
power  delivered  to  any  particular  belt,  shaft,  etc.,  in  the 


ELEMENTARY  PRACTICAL  MECHANICS 


215 


transmission  line.  Transmission  dynamometers  thus  enable 
us  to  determine  the  power  lost  at  any  stage  in  the  power 
transmission,  through  friction,  belt  slip,  etc. 

116b.  Prony  Brakes. — The  simplest  and  most  commonly 
used  form  of  absorption  dynamometer  is  the  prony  brake. 
This  may  be  of  several 
different  forms,  all  of 
which  are  based  upon  the 
same  fundamental  prin- 
ciple. Fig.  154  shows  a 
common  form,  where  A 
indicates  the  belt  pulley 
of  a  motor,  engine,  etc. 
This  is  surrounded  by  the 
brake  band  B,  which  may 


FIG.  154.— Prony  Brake. 


be  tightened  as  desired  by  means  of  the  screw  at  C.  As 
the  pulley  revolves  in  the  direction  of  the  arrow,  friction 
tends  to  carry  the  band  around  also,  but  this  tendency  is 
counteracted  by  a  force  W  acting  on  a  lever  arm  L  attached 
to  the  band. 

If  F  represents  the  force  of  friction, 


and 


The  work  in  ft.-lbs.  absorbed  by  the  brake  in  one  revolu- 
tion is  equal  to  friction  in  Ibs.  times  the  distance  in  feet 
through  which  it  is  overcome,  or 

F  X  circumference  of  pulley. 
And  the  power  absorbed  is  therefore, 

F  X  circumference  of  pulley  X  revolutions  per  minute 
33,000 


216 


ELEMENTARY  PRACTICAL  MECHANICS 


The  friction  brake  evidently  transforms  the  mechanical 
energy  delivered  into  heat.  The  great  amount  of  heat 
thus  generated  usually  makes  it  necessary  to  use  a  cooling 
device  with  the  brake.  Pulleys  with  hollow  rims  which 
may  be  rilled  with  water,  or  hollow  brake  straps  through 
which  a  stream  of  water  is  kept  flowing  are  therefore  often 
used. 

116c.  Transmission  Dynamometers. — The  object  in  all 
transmission  dynamometers  is  to  absorb  as  little  power  as 
possible  in  the  dynamometer  itself  and  to  measure  the 
amount  transmitted.  In  most  common  types,  this  meas- 
urement is  effected  by  determining  either  the  extension 
produced  in  a  system  of  springs  attached  to  the  device,  or 
the  torque  required  to  keep  in  position  a  frame  which  car- 
ries a  system  of  pulleys  on  which  the  driving  belts  run. 


FIG.  155. — Transmission  Dynamometer. 

The  principle  of  the  former  will  be  apparent  from  the 
diagram  of  Fig.  155.  This  shows  a  simple  dynamometer 
coupling  in  which  connection  between  two  parts  of  a 
shaft  is  made  by  the  springs  SSS  instead  of  bolts  or  pins, 
as  in  the  ordinary  rigid  coupling,  plates  A  and  B  are  each 
keyed  to  the  shaft.  As  the  side  B  is  driven,  the  torque  is 
communicated  to  A  through  the  springs  which  are  thereby 


ELEMENTARY  PRACTICAL  MECHANICS  217 

elongated.  If,  now,  the  constant  of  the  springs  is  known, 
also  the  amount  they  are  extended,  and  the  speed  in  revolu- 
tions per  minute,  the  horsepower  transmitted  by  the  shaft 
may  be  readily  computed. 

The  constant  for  the  springs  may  be  found  experimentally 
by  holding  A  and  applying  a  known  torque  to  B  by  means 
of  cords  and  weights.  Suppose,  for  example,  a  torque  of 
25  Ib.-ft.  causes  the  springs  to  extend  2.5  scale  divisions; 
every  scale  division  then  represents  10  Ib.-ft.  of  torque  or 
10X2X7r  =  207r  ft.-lbs.  of  work  per  revolution. 

If,  therefore,  in  an  actual  instance,  the  reading  on  the 
dynamometer  scale  is  10.6  divisions  at  300  r.p.m.,  the 
horsepower  transmitted  is, 


The  practical  difficulty  is,  of  course,  to  transmit  the 
extension  of  the  springs  to  an  index  which  may  be  read 
while  the  shaft  is  rotating,  and  in  the  device  by  which  this 
is  accomplished  lies  the  main  difference  between  the  instru- 
ments in  use. 

In  the  Webber  dynamometer  the  change  in  the  relative 
angular  positions  of  the  plates  operates  levers  which  move 
a  loose  sleeve  on  the  shaft.  This  sleeve  in  turn  operates 
an  index  moving  over  a  graduated  dial  which  gives  the 
horsepower. 

In  the  Emerson  Power  Scale  the  springs  are  replaced  by 
a  weighing  device  somewhat  similar  in  principle  to  a  steel 
yard.  A  and  B  are  connected  by  projecting  studs.  The 
pressure  on  these  studs  is  conve}^ed  by  levers  to  a  collar 
which  in  turn  is  connected  to  a  weighing  lever  where  the 
force  is  measured  by  balancing  weights.  Sudden  motion 
of  the  weighing  lever  is  prevented  by  dash-pots. 

The  preceding  types  are  adapted  to  use  with  a  shaft. 


218  ELEMENTARY  PRACTICAL  MECHANICS 

By  the  addition  of  a  pulley  to  the  shaft  to  which  A  is 
keyed,  they  may  also  be  adapted  to  measure  the  power 
transmitted  by  a  belt.  Two  belts  must  be  used  in  place 
of  one,  the  shaft  or  machine  being  driven  through  the  dy- 
,  _  .  namometer.  The  belt  from  the 

motor  or  engine  drives  B,  and 
B  drives  A  through  the  connect- 
ing springs  or  studs.  A  second 
belt  runs  from  the  extra  pulley 
keyed  to  the  shaft  of  A,  to  the 
machine  to  be  driven.  Compu- 
tation, devices  for  reading,  etc., 

FIG.  156.—  Belt  Dynamometer.  are  tne  same  as  before. 

A  second  type  of  transmission 

dynamometer  for  use  with  a  belt  is  shown  in  Fig.  156  (see 
Church's  "  Mechanics  of  Materials  ").  A  vertical  plate 
carrying  four  pulleys  upon  which  the  driving  belt  runs  is 
balanced  about  a  pivot  at  C.  Power  is  then  turned  on,  and 
the  plate  again  balanced.  If  G  =  weight  to  balance, 


(P  and  Pf  on  right  pass  through  C  and  therefore  have  no 
moment). 

Gb 

Therefore,  P  -  P'  =  —  . 

a 

Work  per  minute  =  (P—  P')v, 

when  v  is  velocity  of  belt  in  feet  per  minute. 

For  a  more  detailed  discussion  of  particular  types  of 
transmission  dynamometers,  the  student  is  referred  to 
engineering  texts,  trade  periodicals,  etc. 

117.  Applications  of  the  Preceding  Principles  to  Simple 
Machines.  —  The  study  of  mechanism  is  regarded  as  entirely 


ELEMENTARY  PRACTICAL  MECHANICS 


219 


outside  the  scope  of  this  book.  The  preceding  discussion, 
together  with  the  solutions  which  follow,  are  intended  to 
familiarize  the  student  with  some  of  the  more  general 
technical  terms  and  computations  which  apply  to  machin- 
ery, and  above  all,  to  enable  him  to  approach  the  formal 
study  of  mechanism  later 
with  a  right  point  of 
view.  A  complete  and 

formal     discussion     of  ijs" 

what  are  generally 
termed  the  "  Simple  Me- 
chanical Principles/ '  has 
been  purposely  avoided. 

118.  The  Winch.— 
Neglecting  friction,  what 
load  can  be  lifted  by  a 
driving  force  of  50  Ibs.  -FIG.  157.— Winch, 

applied  at  right  angles 
to  the  crank  of  the  winch  shown  in  Fig,  157? 

In  one  revolution,  driving  force  moves  through  the  dis- 
tance equal  to  the  circumference  of  the  circle  swept  out  by 
the  crank,  or 


15 
-X2X- 


15 

=  — 7T  ft. 


In  the  same   time  load  is  lifted  a  distance  equal  to  the 


circumference  of 


=  jX7r  =  -  ft. 

a 


Velocity  ratio  therefore  = =5, 


or  driving  force  moves  5  ft.  to  lift  L  I  foot.     Therefore, 
50  X  5 = L  X 1    and    L  =250  Ibs, 


220  ELEMENTARY  PRACTICAL  MECHANICS 

119.  Gear,  Sprocket,  and  Belt  Machines. — In  a  system  of 
gears,  as  shown  in  Fig.  158,  A  has  80  teeth,  B  has  60,  and 
C  has  36.  Diameter  of  drum  Di=3  ins.,  of  I>2  =  9  ins. 
Load  L  =  200  Ibs.  Efficiency  of  system  for  this  load  =80%. 
What  driving  force  will  be  required  at  E? 

Wheel  B  is  here  an  "  idle  wheel/'  and  serves  merely  to 
change  the  direction  of  rotation  of  C.  The  teeth  prevent 


FIG.  158. — Gearing  arranged  for  Laboratory  Test. 

slipping,  hence  linear  speed  of  teeth  on  A,  B,  and  C  must 
be  the  same,  and  the  speed  of  C  will  be  the  same  as  if  A 
were  geared  directly  on  C.  In  one  revolution  of  (7,  E 
moves  the  circumference  of  D2  =  97i:  ins. 

When   C   makes   one   revolution,   the   same   number   of 

36 

teeth  pass  the  point  of  contact  on  A,  or  A  makes  —  of  a 

ou 

revolution.     L  is  therefore  lifted 


ELEMENTARY  PRACTICAL  [MECHANICS 

36     27 


221 


T97T 

Velocity  ratio  =  — -  =  6  § , 
20* 


or  E  moves  6§  ins.  to  raise  L  I  in.     Therefore, 


and 


80 
— =  200X1, 


37.51bs. 


Fig.  159  shows  an  arrangement  for  testing  the  efficiency 
of   transmission   of    the 
chain  drive  of  a  simple 
bicycle. 

The  pedals  have  been 
removed  and  a  wooden 
pulley  having  a  radius 
equal  to  the  length  of 
the  crank  has  been  at- 
tached to  one  crank.  A 
pound  applied  to  the  rim 
of  this  pulley  thus  pro- 
duces the  same  torque 
as  an  equal  force  for- 
merly did  when  applied 
to  the  pedal.  A  resist- 
ing torque  is  applied 
to  the  rear  wheel  by  a 
weight  on  the  cord  run- 
ning on  the  rim,  from  which  the  tire  has  been  removed. 
If  desired,  the  machine  may  now  be  driven  by  the  force  on 


FIG.  159.— Bicycle  Chain  Drive 
arranged  for  Laboratory  Test. 


222  ELEMENTARY  PRACTICAL  MECHANICS 

the  rear  wheel.  This  will  raise  a  greater  weight  on  the 
pulley,  thus  gaining  a  large  "  mechanical  advantage/7  Or  it 
may  be  driven  in  the  usual  way  as  a  "  speed  "  machine  with 
the  force  on  the  pulley  used  as  the  driving  effort. 

A  two-ounce  load  is  hung  from  the  rear  wheel  and  then 
enough  weight  put  on  the  pulley  to  just  raise  this  at  a 
uniform  speed.  Then  weights  are  removed  from  E  until 
the  loadL  lowers  at  the  same  uniform  rate.  Half  the  dif- 
ference between  these  values  (effort  to  raise  arid  effort  to 
lower  the  load)  represents  the  friction  of  the  machine  at  this 
load.  The  driving  effort  for  a  frictionless  machine  can  be 
found  by  taking  half  their  sum.  The  input,  output,  and 
efficiency  are  computed  and  curves  plotted  to  show  the  rela- 
tion of  efficiency  and  driving  effort  to  load  raised.  Also  on 
the  same  sheet  the  driving  efforts  for  a  frictionless  machine 
are  plotted  against  the  loads.  The  increase  of  friction 
with  added  loads  is  brought  out  by  the  divergence  of  the 
two  "  effort  "  curves.  The  displacement  ratio  is  computed 
by  measuring  the  diameters  of  the  pulley  and  wheel,  and 
counting  the  number  of  teeth  on  each  sprocket.  The  com- 
puted value  is  then  tested  by  actual  measurements  of  the 
distances  moved  over  simultaneously  by  effort  and  load. 

Below  is  the  report  of  two  students,  showing  results  of 
a  test  with  this  apparatus: 

THE    CHAIN   DRIVE 
DATA 

Diameter  of  driver,  12.0  ins. 
Diameter  of  follower,  25.0  ins. 
Number  of  teeth  on  front  sprocket,  18. 
Number  of  teeth  on  rear  sprocket,  7. 

driver        driver'       12      7     ^  ., 

Displacement  ratio  =  7— ~ Xf— n ?— TTF  X  -r^ =0. 187. 

follower     follower'     25     18 


ELEMENTARY  PRACTICAL  MECHANICS 


223 


Displacement  ratio  = 

Distance  moved  by  effort,  6.7  ins.  _~ 
Distance  moved  by  load,  36.0  ins. 

Displacement  ratio  (computed)  =0.187. 

Displacement  ratio  (checked)     =0.19. 


DATA. 


COMPUTED  VALUES. 


Load  in 
Lbs. 

Effort  to 
Lift 
Loads. 
Lbs. 

Effort  to 
Lower 
Load. 
Lbs. 

Effort  for 
Friction- 
less 
Machine. 

Effort  to 
Overcome 
Friction. 

Input, 
Ft.  Lbs. 

Output. 
Ft.Lbs. 

Per  Cent 
Effi- 
ciency 

.13 

.94 

.63 

.78 

.16 

.18 

.13 

72 

.25 

1.59 

1.28 

1.44 

.16 

.30 

.25 

83 

.50 

2.90 

2.56 

2.73 

.17 

.551 

.50 

91 

1.00 

5.60 

5.18 

5.39 

.21 

1.06 

1.00 

94 

1.50 

8.38 

7.75 

8.02 

.32 

1.59 

1.50 

94 

2.00 

11.0 

10.3 

10.7 

.35 

2.09 

2.00 

96 

2.75 

15.0 

14.1 

14.6 

.45 

2.85 

2.75 

97 

3.75 

20.5 

19.3 

19.9 

.60 

3.89 

3.75 

97 

5.75 

31.2 

29.9 

30.6 

.65 

5.93 

5.75 

97 

6.88 

37.4 

36.0 

36.7 

.70 

7.10 

6.88 

97 

between    driving   efforts   and   loads 
driving  effort  is  directly  pro- 


m 


The  curve  plotted 
shows  that  the  increase 
portional  to  any  increase 
in  load.  The  curve  does 
not  pass  through  the  ori- 
gin because  it  requires 
some  effort  to  drive  the 
machine  unloaded;  the 
" y "intercept  shows  this 
effort  to  be  \  Ib. 

The  curve  between 
loads  and  efforts  for  a 
frictionless  machine 
shows  that  these  efforts 
are  directly  proportional 
to  the  loads  themselves. 
This  curve  passes  through  the  origin  because  no  effort  would 
be  required  to  drive  an  unloaded,  frictionless  machine," 


L,B8 


FIG.  160. 


224  ELEMENTARY  PRACTICAL  MECHANICS 

The  curve  between  efficiency  and  loads  shows  that, 
within  the  limits  of  the  experiment,  the  efficiency  at  first 
increases  as  the  load  increases,  until,  at  a  load  of  2f  Ibs., 
the  efficiency  becomes  practically  constant  at  97%. 

120.  The  Jack  Screw. — A  jack  screw  is  turned  by  a 
lever  1J  ft.  long  from  center  of  screw.  There  are  six 
threads  to  the  inch  on  the  screw.  A  force  of  100  Ibs., 


FIG.  161. — Jack  Screw  equipped  for  Laboratory  Test. 

applied  at  the  end  of  the  lever  and  at  right  angles  to  it, 
will  lift  25,400  Ibs.  What  is  th6  efficiency  of  the  jack  at 
this  load? 

Each  revolution  of  the  screw,  the  driving  force  moves  2  X 

3     22 

7rX-=-=9.42  ft.,  and  the   load  is  lifted  distance  between 

^       7 

threads,  or  J  in.  —  -fa  ft.     Therefore, 

100  X  9.42  X  efficiency  =  25,400  X  -fa, 
and  Efficiency  =  37.6%. 


ELEMENTARY  PRACTICAL  MECHANICS  225 

The  apparatus  shown  in  Fig.  161  is  arranged  for  the 
test  of  a  small  jack-screw.  No  attempt  is  made  to  intro- 
duce any  refinements  into  either  the  apparatus  or  the 
methods  of  making  readings,  but  the  purpose  has  been  to 
retain,  as  nearly  as  possible,  the  actual  working  condi- 
tions obtaining  in  practice. 

The  jack-screw  is  a  small  one  having  a  head  upon  the 
screw  spindle,  turning  on  roller  bearings.  The  screw  has  a 
pitch  of  f  in.  The  base  rests  upon  a  heavy  plank  suitably 
supported.  To  the  head  of  the  screw  is  fastened  a  grooved 
wooden  wheel,  and  the  screw  is  turned  by  spring  balances 
attached  to  cords  wrapped  about  this  wheel. 

Across  the  head  rests  a  round  steel  rod,  and  upon  this 
rod  rests  a  substantial  wooden  lever  about  9  ft.  long.  This 
lever  has  one  end  fastened  down  to  the  floor  by  an  iron  rod 
provided  with  a  thread  and  nuts  at  the  top,  so  that,  as  the 
screw  is  used  at  different  portions  of  its  length,  the  lever 
may  always  be  kept  horizontal.  A  scale  is  attached  to  the 
lever,  so  that  the  distance  of  joint  A  from  the  lever  ful- 
crum at  B  may  be  read  easily.  (See  Fig.  12.)  By  shifting 
the  jack-screw  along  the  floor  the  ratio  of  the  lever  arms 
may  be  varied  within  certain  limits. 

The  weight  of  the  screw,  head  and  wheel  are  determined 
and  also  that  of  the  lever.  The  center  of  gravity  of  the 
lever  is  found  by  balancing  it  across  a  knife  edge  and  its 
position  is  marked  on  the  lever.  At  C  is  a  clevis  and 
scale-pan  for  supporting  the  variable  loads  as  desired. 

The  methods  used  in  calculation  are  as  follows  (see 
Fig.  12): 

To  find  the  vertical  pressure  on  the  screw  spindle  for 
any  weight,  L,  at  C. 
Pressure  =  (weight  of  wheel,  etc.) 


weight  of  leverX-fLx 


226  ELEMENTARY  PRACTICAL  MECHANICS 

In  the  test  reported  below  this  gives: 


Pressure  at  J5  =1 


18   /       18 


Velocity  Ratio:  In  one  revolution  of  the  wheel  the 
resistance  at  B  is  overcome  through  a  distance  equal  to  the 
pitch  of  the  screw,  or  f  in. 

The  effort,  applied  to  the  rim  of  the  wheel,  will  in  one 
revolution  move  a  distance  equal  to  the  circumference  of 
the  wheel,  or  20X3.14  =  62.8  ins.  Therefore  the  velocity 
ratio  is 

^=167.4. 


This  is  equivalent  to  saying  that  for  one  foot  moved  by 
the  weight  at  B  the  effort  on  rim  of  wheel  must  move 
167.4  ft.  These  figures  are  the  ones  used  in  the  calcula- 
tion of  input  and  output  in  the  table  below. 

To  find  the  force  of  friction  and  the  effort  that  would 
be  required  if  there  were  no  friction,  we  proceed  as 
follows : 

Let  E  =  effort  with  no  friction; 
Fr  =  friction; 

Si  =  effort  for  a  rising  load  at  C; 
$2  =effort  for  a  descending  load  at  C. 

NOTE. — All  these  four  quantities  are  here  understood  to  be  the 
force  in  pounds  at  the  rim  of  the  wheel. 

Then  E  must  be  increased  by  Fr-  to  give  the  force  to  lift 
the  load,  or 

(a)  Si  ' 


ELEMENTARY  PRACTICAL  MECHANICS 


227 


The  machine  will  not  run  back  of  itself,  but  a  force  must 
be  applied  to  turn  it  back;  this  force  is  $2.     Then 


or 


(W 

Adding  (a)  and 


Si '+S2 


-Fr. 


Subtracting  (b)  from  (a) 

Si—  S2 


Since  the  velocity  ratio  is  167.4,  Ex  167.4  should. give 
the  pressure  at  B. 

In  the  results  of  student  tests  given  below,  the  loads  were 
determined  by  weighing  L  on  a  platform  balance.  The 
efforts  at  rim  of  wheel  were  found  by  spring  balances,  a 
four  pound  balance  reading  to  ounces  for  the  lower  reading 
and  a  larger  balance  for  the  higher  ones. 


^ 

Efforts  at  Rim  of  Wheel. 

^" 

i 

ti 

37 

^+ 

T3 

S 

. 

CO 

rH 

X 

X 

1 

i3 

Jj  i> 

IN 

§1^ 

«2-| 

^1 

OQjg 

*l 

«« 

?8 

tfa 

^^ 

|JB 

Is' 

p' 

a  Is 

.2  a 

H 

& 

£'~c 

W 

£ 

fl 

0 

H 

0 

72 

1.19 

.25 

.47 

.72 

199 

72 

36.2 

8.29 

117 

1.81 

.38 

.71 

1.09 

302 

117 

38.8 

19.1 

178 

2.75 

.63 

1.06 

1.69 

460 

178 

•38.8 

40.6 

229 

4.25 

.75 

1.75 

2.50 

710 

299 

42.0 

60.6 

406 

6.00 

.94 

2.53 

3.47 

1000 

406 

40.6 

80.6 

515 

7.50 

1.75 

2.88 

4.63 

1250 

515 

41.2 

102.1 

632 

9.50 

2.00 

3.75 

5.75 

1590 

632 

39.8 

122.3 

742 

11.00 

2.25 

4.37 

6.62 

1840 

742 

40.3 

132.2 

800 

12.00 

2.50 

4.75 

7.25 

2000 

800 

40.0 

228 


ELEMENTARY  PRACTICAL  MECHANICS 


It  is  apparent  from  the  table  that  the  efficiency  is  nearly 
the  same  for  all  the  loads  used,  though  there  is  a  slight 

increase  for  the  larger 
loads.  If  still  larger  loads 
were  used  there  might  be 
more  variation  in  the 
efficiency. 

The  results  of  this  test 
are    also    shown   in  the 


in 

plotted  chart,  see  Fig. 
162,  where  total  load,  P, 
and  effort  for  a  rising 
load,  S,  are  plotted  and 
also  total  load  and  fric- 
tion. It  appears  here 
that  both  the  effort  and 

the  friction  are  proportional  to  the  total  load  within  the 

range  of  loads  used. 


400 
Load  in  Pounds 

FIG.  162. 


121.  The  Block  and  Tackle. — In  the  pulley  blocks  shown 
in  Fig.  163,  suppose  L  were  lifted  1  in.,  then  each  of  the 
five  cords  supporting  L  would  be  shortened  1  in.  and  to 
keep  the  rope  tight  E  must  move  5  ins. 

Velocity   ratio   therefore  =  5  =  number   of   parts   of   rope 
supporting  movable  block. 

122.  The  Chain  Hoist.— The  Weston  differential  pulley, 
used  for  hoisting  loads,  consists  of  two  sheaves  of  different 
diameters  in  the  upper  block,  rigidly  fastened  together,  and 
one   sheave  in  the   lower  block.     An  endless    chain  runs 
over  these  blocks,  the  rims  of  the  sheaves  being  formed  so 
as  to  prevent  the  chain  from  slipping  upon  them.     The  ar- 
rangement is  shown  in  Fig.  164.     Suppose  R  be  radius  of 
larger  sheave,  r  radius  of  smaller.     To  hoist  load  L,  the 
driving  force  E  is  applied  to  chain   coming   from   larger 
sheave  as  shown.     Suppose  upper  pair  of  sheaves  make 
one  revolution.     Then  E  moves  a  distance  2xR,  side  B  of 
chain  is  raised  a  distance  2/Tjft,  but  side  A  is  lowered  at  the 


ELEMENTARY  PRACTICAL    MECHANICS 


229 


same  time  a  distance  27rr,  as  it  unwinds  from  the  smaller 
sheaves.  The  loop  of  chain  AB  is  actually  shortened, 
therefore,  a  distance  2xR  —  27rr  =  27r(E-r).  The  load  L  is 
raised  one-half  of  this  distance,  or  n(R-r). 


Velocity  ratio  = 


2R 


n(R-r)     R-r 


FIG.  163.— Block  and 
Tackle. 


FIG.  164. — Weston  Differential 
Pulley. 


By  making  the  sheaves  nearly  the  same  size  (i.e.,  R— r 
small),  this  may  be  increased. 

The  energy  equation  for  the  apparatus  would  be, 

EX2RX  efficiency = L  X  (#  -  r*) . 
As  used  this  pulley  does  not  run  down  when  force  E  is 


230 


ELEMENTARY  PRACTICAL  MECHANICS 


removed.    The  tendency  to  turn  clockwise,  as  will  be  seen 

L 

from,  the  figure,  is  —  Xr.     The  counter-clockwise  moment 

tU 

L 

Is    -XR.      The    counter-clockwise   moment   therefore   ex- 


ceeds the  clockwise  by  the  difference,  or 

L 

— (12— r).     Friction  of  the  apparatus  (upper 

4* 

and  lower  blocks)  must  exceed  this  and 
then  the  load  will  not  run  down. 

The  chain  hoist  furnishes  an  excellent 
medium  for  a  laboratory  study  of  a  "  non- 
re  versible"  machine.  An  arrangement  of 
apparatus  for  such  a  test  of  a  quarter- 
ton  hoist  with  8-ft.  lift,  is  shown  in  Fig. 
165. 

The  block  is  fastened  to  an  I-beam. 
The  load  consists  of  an  iron  cradle  and 
nine  50-lb.  weights  which  may  be  added 
one  at  a  time  in  order  to  test  the  ma- 
chine under  different  loads  from  50  to 
550  Ibs. 

The  procedure  is  as  follows:  The  di- 
ameters of  the  differential  pulleys  are  first 
measured  and  the  displacement  ratio 

2R 


165.  — Quar- 


ter-ton Chain    computed    from    the     formula 

TT      •     i       T  J      J  A 


Hoist  Loaded. 


R-r 


where  R  is  radius  of  large  pulley  and  r 
the  radius  of  the  smaller  one.  This  is  then  checked  by 
measuring  the  length  of  chain  running  over  large  pulley, 
which  has  to  be  displaced  in  order  to  raise  the  load  one 
foot. 
The  effort  required  to  raise  a  given  load  is  measured 


ELEMENTARY  PRACTICAL  MECHANICS 


231 


by  pulling  on  a  spring  balance  attached  to  the  chain.  As 
the  machine  will  not  run  backwards,  a  negative  effort, 
i.e.,  an  effort  on  the  other  side  of  same  pulley,  must  be 
applied  to  lower  the  load.-  The  amount  of  negative  effort 
is  determined  in  the  same  way  as  the  effort  required  to 
raise  the  load.  Half  the  algebraic  difference  of  these  two 
quantities  is  approximately  the  value  of  the  friction  at 
that  load.  The  output  is 
computed  by  assuming 
load  to  be  lifted  1  ft., 
thus:  Output  =  load  X 
1  ft.-lbs.  The  input  then 
equals  the  product  of  the 


60  .60 

^       J 
§  iO-llO 

&   I 

H 

displacement  ratio,  by  the    §    | 
effort  to   raise  the    load. 
-By    dividing    output    by 


100  200 

Load  in  Pounds 


.  .  .      FIG.  166.— Relation  of  Efficiency,  Dnv- 

mput,    the     efficiency    is      ing  Effort>  and  Function  to  Load  in 

found  at  the  given  load.       Chain  Hoist. 
This  is  repeated  for  loads 

increasing  by  50-lb.  steps  until  a  load  of  about  550  Ibs.  is 
reached.     Curves  are  then  plotted,  showing  relation  between 
load  and  effort,  load  and  friction,  and  load  and  efficiency. 
Below  is  a  report  of  two  students  on  this  apparatus. 


THE    CHAIN    HOIST 

Radius  large  pulley  #  =  4.5  ins. 
Radius  small  pulley  r  =  4.0  ins. 

2R        2X4.5      1Q 
Displacement  ratio  =  -H"~  ^  4  5  —4  Q  =  y* 

Check  of  displacement  ratio  = 
distance  effort  moves  _  72  jns. 
distance  load  moves       4  ins. 

The  curves,  Fig.   166,  plotted  from  this  data  show,  an 
Intercept  on  the  Y-axis  of  \  Ib.     This  is  the  driving  effort 


232 


ELEMENTARY  PRACTICAL  MECHANICS 


to  overcome  friction  with  no  load.  The  efficiency  increases 
to  about  30%  at  75  Ibs.  load  and  then  remains  practically 
constant  within  the  limits  of  the  test. 


DATA. 


COMPUTED  RESUTS. 


Load 
in 
Lbs. 

Effort  to 
Raise  Load, 
Lbs. 

Effort  to 
Lower 
Load,  Lbs. 

Friction, 
Lbs. 

Input, 
Ft.-Lbs. 

Output, 
Ft.-Lbs. 

Efficiency 
Per  Cent. 

61 

13 

4 

8.5 

234 

61 

26 

87 

16 

6 

11 

288 

87 

30 

108 

20 

7 

13.5 

360 

108 

30 

137 

25 

9 

17 

450 

137 

30 

165 

30 

11 

20.5 

540 

165 

31 

191 

36 

13 

24.5 

648 

191 

30 

216 

41 

15 

38 

738 

216 

29 

240 

46 

16 

31 

828 

240 

29 

286 

52 

20 

36 

936 

286 

31 

359 

60 

24 

42 

1080 

359 

33 

123.  Energy  in  Starting  and  Stopping  Machines. — In  the 

act  of  starting  any  machine  under  a  given  load  a  greater 
supply  of  energy  must  be  furnished  than  is  required  to 
maintain  the  motion  at  a  uniform  speed  when  started, 
because  every  moving  part  must  be  accelerated  from  rest 
up  to  some  velocity.  Similarly  a  machine  in  motion  has 


ft.-lbs. 


/WV2 
stored  in  its  parts  a  supply  of  kinetic  energy  ( 

\     Q 

which  will  be  available  to  do  some  work  even  after  the 
actual  driving  power  ceases  to  act.  (If  the  force  of  fric- 
tion is  large  the  useful  work  done  may  be  so  small  as  to 
be  not  very  apparent.)  If  the  throttle  of  a  locomotive  be 
closed  after  the  locomotive  is  set  in  motion,  the  kinetic 
energy  of  the  locomotive  and  train  may  suffice  to  carry 
them  some  distance  along  the  track  against  the  resistance 
of  friction  and  perhaps  of  gravity  also  on  an  up  grade. 
Furthermore  it  should  be  noted  that  the  kinetic  energy  of 


ELEMENTARY  PRACTICAL  MECHANICS  233 

all  the  rotating  wheels  is  available  as  well  as  the  energy  of 
the  train  along  the  rails.  All  this  kinetic  energy  had  to 
be  supplied  to  the  train  either  by  the  engine  or  gravity  (on 
a  down  grade),  or  both,  in  the  act  of  starting  the  train 
from  rest,  and  remains  stored  until  in  stopping  it  is  finally 
dissipated,  chiefly  as  heat. 


PROBLEMS 

1.  A  safe  weighing  5  tons  is  to  be  loaded  on  a  truck 
4  ft.  high  by  means  of  planks  18  ft.  long.     If  it  requires 
200  Ibs.  to  overcome  friction  of  the  safe  on  the  planks,  find 
least  force  which  will  be  necessary. 

2.  A  safety  valve  is  arranged  as  in  Fig.  167.     Weight  of 
AB  =  12  Ibs.,  weight  acts  10 

ins.  from  A.  Weight  of  valve 
V  =  8  Ibs.  Diameter  of  valve 
3  ins.  What  effective  steam 
pressure  per  square  inch  will 
just  open  the  valve? 

3.  In   a   letter  press,  the 

diameter  of  the  wheel  is  14  ins.,  pitch  of  thread  is  ^  in. 
Find  pressure  applied  by  a  pull  of  50  Ibs.  tangent  to  the 
wheel  if  the  efficiency  of  the  press  is  30%. 

4.  In  an  ordinary  "  jack-screw,"  screw   is  turned  by  a 
lever  2  ft.  long    (from  center  of  screw).     If  there  are  4 
threads  to  the  inch,  find  force  required  on  the  end  of  the 
lever  to  lift  10,000  Ibs.  when  65%  of  the  total  effort  is 
required  to  overcome  friction. 

5.  If  an  effort  of  100  Ibs.  acting  upon  a  machine  moves 
with  a  velocity  of  10  ft.  per  second: 

(a)  How  great  a  load  can  it  give  a  velocity  of  125 
ft.  per  second? 

(6)  With  what  velocity  can  it  move  a  load  of  200 
Ibs.? 


234  ELEMENTARY  PRACTICAL  MECHANICS 

6.  The  pilot  wheel  of  a  boat  is  3  ft.  in  diameter;    the 
axle  is  6  ins.     The  resistance  of  the  rudder  is  180  Ibs.  at 
the  axle.     What  effort  applied  to  the  wheel  will  move  the 
rudder? 

7.  Four  men  are  hoisting  an  anchor  weighing  one  ton. 
Barrel  of  capstan  is  8  ins.  diameter,  length  of  hand  spikes 

3  ft.  4  in.    (from  center  of  barrel).     How  great  pressure 
must  each  man  exert?     What  is  the  velocity  ratio? 

8.  In  moving  a  building,  the  horse  is  attached  to  a  lever 
7  ft.  long,  acting  on  a  capstan  barrel  11  ins.  diameter;    on 
the  barrel  winds  a  rope  of  a  system  of  2  fixed  (near  capstan) 
and  3  movable  pulleys.     What  force  will  be  exerted  when 
the  horse  pulls  500  Ibs.,  allowing  50%  loss  from  friction? 

9.  In   a  simple   wire-testing   machine  the   displacement 
ratio  is  480.     An  effort  of  1.6  Ibs.  will  support  a  load  of 
32  Ibs.     Compute: 

(a)  Input  and  output  in  moving  load  2  ins. 

(&)  Efficiency  of  machine  at  this  load. 

(c)   Effort  expended  in  overcoming  friction. 

10.  A  railway  engine  hauls  a  train  on  a  level  track  at  a 
speed  of  30  miles  per  hour  against  an  average  resistance  of 
4500  Ibs.     It  burns  760  Ibs.  of  coal  per  hour.     If  the  com- 
bustion of  each  pound  of  coal  supplies  10,000,000  ft.-lbs. 
of  energy,   compute  input,   output,  and  efficiency  of  the 
engine. 

11.  Diameter  of  wheel  A,  Fig.  168,  is  16  ins.; 

"       "    5,  10  ins.; 

"  drum  (7,  12   " 

"       "    D,    3   " 

If  5  Ibs.  are  required  at  E  to  overcome  friction  when  L 
=  90  Ibs.,  what  value  of  E  will  raise  L  uniformly?  What 
is  efficiency  of  apparatus  at  this  load? 

12.  By  experiment  with  a  Weston  pulley  it  was  found 
that  a  pull  of  15  Ibs.  on  the  leading  side  was  required  to 
lift  60  Ibs.,  including  lower  pulley.     The  larger  pulley  was 

4  ins.   diameter,  the  smaller,   3J  ins.     Find  efficiency  at 
this  load.     (See  Fig.  164.) 


ELEMENTARY  PRACTICAL  MECHANICS 


235 


13.  In  the  worm  and  worm-wheel,  Fig.  169,  wheel  A  has 
98  teeth,  and  drum  B  is  8  ins.  diameter.  The  worm  is 
turned  by  a  handle  18  ins.  long.  What  is  the  velocity  ratio? 


FIG.  168. 


FIG.  169. 


14.  In  an  arrangement  of  gears,  as  in  Fig.  158, 

A  has  72  teeth, 

B    "    60     " 

C   "    32     " 

Drum  for  E,  6  ins.  diameter,  for  L,  4  ins.  diameter.  Find 
efficiency  of  the  apparatus  and  the  force  required  at  E  to 
overcome  friction  when  a  force  of  30  Ibs.  at  E  will  raise  a 
load  85  Ibs. 

15.  A  24-in.  pulley  on  an  engine  shaft  makes  210  revo- 
lutions per  minute  and  drives  by  belt  a  30-in.  pulley  on  a 
second  shaft.     This  shaft  carries  a  worm  which  gears  with 
a  wheel  on  a  third  shaft.     If  this  wheel  has  108  teeth, 
compute  speed  of  third  shaft. 

16.  An  engine  shaft  running  120  revolutions  per  minute 
carries  a  pulley  42  ins.  diameter,  which  drives  by  belt  a 
pulley  27  ins.   diameter  on  a  line  shaft.     The  line  shaft 
carries  a  second  pulley  48  ins.  diameter,  which  connects 
by  belt  to  a  pulley  20  ins.  diameter  on  a  counter  shaft. 
Speed  of  the  counter  shaft? 

What  size  second  pulley  must  be  used  on  the  counter 
shaft  to  connect  by  belt  to  a  dynamo  armature  having  an 
8-in.  pulley,  if  speed  of  the  dynamo  is  to  be  1400  revolutions 
per  minute? 


CHAPTER  XII 
ELASTICITY 

124.  Elastic  Material.    Experiment  with  a  Steel  Wire. — 

In  our  study  of  bodies  in  equilibrium  and  in  motion  we 
have  assumed,  in  every  instance,  that  the  bodies  were 
"  rigid."  As  a  matter  of  fact,  the  application  of  external 
forces  always  produces  a  change  in  the  form  of  a  body:  the 
amount  and  character  of  such  change  being  dependent 
upon  the  amount  and  arrangement  of  the  applied  forces, 
and  upon  the  material  of  which  the  body  is  composed. 

As  a  simple  illustration,  consider  a  wire,  as  A B  Fig.  170, 
attached  to  a  rigid  support  at  A  and  carrying  a  scale-pan 
at  B.  By  adding  suitable  weights  to  the  scale-pan,  we 
may  subject  the  wire  to  a  pull  of  any  desired  amount.  To 
determine  the  effect,  suppose  we  place  the  wire  in  slots  in 
the  guides  GG  to  prevent  swinging,  and  then  fasten  a 
thread  attached  to  the  short  arm  of  a  light  pointer  P, 
moving  freely  about  an  axis  at  Z>,  to  a  point  on  the  wire 
by  means  of  a  small  spring  clip  C  *  clasped  about  the 
wire.  If  the  wire  stretches,  the  clip  C  will  be  carried 
downward,  the  pointer  will  be  rotated  about  the  axis,  and 
the  long  arm  will  be  moved  along  the  steel  scale  S  by  an 

length  of  long  arm 

amount  equal  to  , — — • X  distance  C  moves. 

length  of  short  arm 

*  The  ordinary  triangular  spring  wire  clips  used  for  fastening 
together  loose  sheets  of  paper  have  proven  perfectly  satisfactory 
for  this  purpose. 

236 


ELEMENTARY  PRACTICAL  MECHANICS 


237 


To  correct  for  errors  in  our  experiment  arising  from  the 
sagging  of  the  support  at  A,  etc.,  a  second  pointer  should 
be  attached  in  the  same  manner  to  a  point  on  the  wire 
near  A.  The  length  of  wire  under  investigation  is  thus 
limited  to  the  portion  between  the  clips  C  and  C",  and  the 
actual  stretch  produced  in  this  length  by  any  given  load 
at  L  is  measured  by  the  amount  that 
the  lower  clip  moves  more  than  the  upper. 
By  this  arrangement,  a  wire  8  ft.  or 
more  in  length  may  be  used  conveni- 
ently in  an  ordinary  room.  With  point- 
ers 10  ins.  from  scale  to  axis  D  and 
1  in.  from  D  to  center  of  hole  where 
thread  from  clip  is  attached,  and  scales 
on  which  the  positions  of  the  pointers 
may  be  read  to  j^th  inch,  the  stretch 
in  8  ft.  of  wire  can  be  measured  to  about 
— ^th  inch,  or  about  10Q1QOQth  inch  in 
each  inch  of  length.  Closer  readings 
should  not  be  attempted. 

In  performing  our  experiment,  suppose 
we  first  apply  a  load  of  10  or  20  Ibs.  to 
remove  any  kinks  from  the  wire,  and 
then  having  first  set  each  pointer  near 
the  bottom  of  its  scale  and  noted  its 
position,  suppose  we  add  weights  in  5-lb. 
steps  to  the  scale-pan,  and  read  the  positions  of  the  point- 
ers after  each  increase  in  load.  Since  we  are  concerned  for 
the  moment  only  with  the  effect  of  a  given  increase  in 
load,  we  may  disregard  the  effect  of  the  "  straightening 
load"  and  assume  that  we  start  from  zero  load. 

Readings  for  a  piano  steel  wire  0.0348  in.  diameter  and 
90  ins.  long  between  clips  are  shown  in  the  following  table: 


238 


ELEMENTARY  PRACTICAL  MECHANICS 


Added 
Load. 
Lbs. 

Position  of  Pointers. 

Total 
Apparent 
Stretch. 

Total 
Actual 
Stretch. 

Stretch 
Per  Inch 
(Strain.) 

Load  in 
Lbs.  Per 
Sq.  Inch 

(Stress). 

Upper. 

Lower. 

0 

1.14 

0.16 

5 

1.19 

0.37 

0.16 

0.016 

0.00018 

5,300 

10 

1.21 

0.56 

0.33 

0.033 

0.00037 

10,500 

15 

1.23 

0.73 

0.48 

0.048 

0.00053 

15,900 

20 

1.24 

0.90 

0.64 

0.064 

0.00071 

21,200 

25 

1.25 

1.06 

0.79 

0.079 

0.00088 

26,500 

0 

1.23 

0.25 

0.00 

0.000 

0.00000 

00,000 

From  these  figures,  it  is  apparent  that  the  wire  when 
supporting  a  scale  pan  and  weights  was  longer  than  the 
same  wire  with  no  load;  in  other  words,  a  pull  caused  the 
wire  to  stretch.  Also,  that  the  wire  continued  to  stretch 
more  and  more  as  the  load  increased  but  immediately 
returned  to  its  original  length  when  the  stretching  load  was 
removed.  This  is  the  characteristic  property  of  elastic 
material;  therefore  we  may  say  that  the  steel  wire  was 
elastic  under  the  conditions  present  in  the  experiment. 
(The  permanent  change  in  the  positions  of  the  pointers  was 
due  to  sagging  of  the  support,  stretch  in  the  wire  above 
the  upper  pointer,  etc.,  as  was  evidenced  by  the  fact  that 
both  pointers  remained  displaced  by  the  same  amount.) 

DEFINITION. — ELASTICITY  may  be  defined  as  the  property 
of  bodies  by  virtue  of  which  they  resume  their  original  form 
when  the  external  forces  acting  on  them  are  removed. 

125.  Stress. — When  a  load  is  placed  in  the  scale  pan  of 
the  apparalus,  Fig.  170,  a  pull  equal  to  the  total  weight  of 
scale  pan  and  contents  is  applied  to  every  section,  across 
the  wire  tending  to  separate  the  particles  of  steel  along  this 
section  from  those  next  above  them.  This  result  is  op- 
posed by  the  cohesion  between  the  pairs  of  particles,  and 
thus  the  pull  is  transmitted  through  the  wire  to  the  sup- 


ELEMENTARY  PRACTICAL  MECHANICS  239 

port.  The  total  pull  at  a  section  is  distributed  over  all  the 
pairs  of  particles  along  that  section.  If  we  divide  the  total 
pull  by  the  area  of  section  in  square  inches,  we  shall  have 
the  pull  per  square  inch  to  which  the  material  is  subjected. 
This  is  called  the  STRESS. 

total  force 
stress  =jorce  per  square  inch  = 


area  of  section 
Thus  when  the  total  pull  is  10  Ibs.,  the  tensile  stress  in  the 


the  cross  section  of  the  wire  is  not  uniform,  the  greatest 
stress  occurs  where  the  diameter  is  least. 

A  stress  may  be  tensile,  as  in  the  preceding  illustration; 
compressive,  as  in  columns  and  struts  supporting  a  load,  in 
which  case  the  particles  are  pushed  closer  together  tending 
to  crush  the  material;  or  shear,  in  which  one  portion  of 
material  tends  to  slide  past  another,  as  in  the  relative 
motion  of  the  blades  of  a  pair  of  shears.  A  shear  stress  is 
exerted  in  punching  holes  in  a  plate,  in  bending  beams, 
etc.  Shear  stress  will  be  considered  in  more  detail  in 
Article  132. 

126.  Strain.  —  When  stress  is  applied  to  a  body,  the  latter 
undergoes  change  of  form;  i.e.,  is  stretched,  compressed, 
bent,  twisted,  etc.  The  fraction  of  its  original  size  by  which 
the  body  changes,  i.e.,  its  change  of  form  per  unit  of  original 
size,  is  called  the  STRAIN. 

total  change  in  size 
Strain  =  -  —     i  —  :  -  • 
original  size 

Thus  when  supporting  a  weight  of  10  Ibs.,  the  steel  wire 
originally  90  ins.  long  was  found  to  have  increased  in  length 


240  ELEMENTARY  PRACTICAL  MECHANICS 

0.033  ins.     The  total  stretch  was  therefore  0.033  ins.,  and 

If  a  column  10.00 


0.033 

the  strain =  0.00037  ins.  per  inch. 

90 


ins.  long  shortens  under  compression  until  only  9.98  ins.  long, 
the  change  in  length  is  0.02  in.  and  the  strain  is  —  -  —  =  0.002 

ins.  per  inch. 

127.  Hooke's  Law.  —  If  we  plot  the  curve  between  the 

last  two  columns  of  the  pre- 
ceding table  of  data,  we  shall 
obtain  the  curve  of  Fig.  171. 
This  is  a  straight  line  pass- 
ing through  the  origin,  thus 
showing  that  for  the  range 
of  stress  to  which  the  wire 
was  subjected,  stretch  was 
directly  proportional  to  load. 
This  illustrates  an  important 
general  law,  for  all  elastic 

FlG,  ^--C^ve  showing  relation    material  whether  under  ten- 
of   stretch  to  load  for  a  piano 

steel  wire.  sl^e>  compressive,    or    shear 

stress.     In  its  general  form, 

this  law  may  be  expressed:   "  Strain  is  directly  proportional 

to  stress  "  (Hooke's  Law).* 

128.  Modulus  of  Elasticity.  —  The  modulus  of  elasticity,  or 
Young's  modulus,  for  a  material  (sometimes  called  also  its 

coefficient  of  elasticity)  is  expressed  by  the  ratio 


20000 
10000 

2 

/ 

/ 

/ 

X 

/ 

« 

/ 

/ 

PIANO 

STEEL  W 

RE.034J 

DIAM. 

.00010 
Stretch  in  inches  per  in. 


. 
strain 


*  We  may  now  see  why  all  elastic  vibrations  are  simple  harmonic 
At  the  position  of  extreme  displacement,  i.e.,  greatest  strain,  the 
returning  force  is  greatest,  hence  here  also  the  acceleration  is  a 
maximum  and  toward  the  center.  At  the  center,  stress,  and  there- 
fore acceleration,  is  zero. 


ELEMENTARY  PRACTICAL  MECHANICS  241 

stress     Ibs.  per  sq.in. 

Modulus  oi  elasticity  for  tension,  E— —  =  — : — • 

strain     stretch  per  in. 

21  200 
For  our  piano  steel  wire,  modulus  of  elasticity,  E  =  — 

0.00071 

=30,000,000  nearly,  which  is  practically  the  accepted  value 
for  most  grades  of  steel. 

The  modulus  of  elasticity  for  compression  may  be  found  in 
a  similar  way  for  short  pieces  or  for  longer  ones  which  are 
prevented  from  bending.  The  piece  is  subjected  to  a 
known  load  tending  to  shorten  it,  and  the  amount  of  de- 
crease in  length  is  measured.  Stress  then  equals  load^- 
area  of  cross  section;  strain  equals  amount  the  piece  is 
shortened -f- original  length  of  piece.  The  modulus  of  elas- 
ticity for  compression  has  practically  the  same  value  as 

that  for  tension,  i.e.,  compressive  strain  = X  com- 

30,000,000 

pressive  stress. 

Long  columns,  if  not  supported  laterally,  yield  by  bend- 
ing. In  such  cases,  as  also  the  deflection  of  beams  under 
transverse  load,  the  stress  is  not  uniform  over  the  section. 
The  computation  for  stress  and  for  modulus  of  elasticity  for 
bending  is  more  complicated,  and  will  be  considered  later. 

129.  Elastic  Limit. — We  have  seen  that  for  the  stress 
applied  in  our  experiment  with  the  piano  steel  wire,  strain 
was  directly  proportional  to  the  stress  producing  or  accom- 
panying it,  and  that  under  these  conditions  the  steel  was 
elastic  and  therefore  recovered  immediately  from  strain 
when  the  pull  was  removed.  If,  however,  we  had  con- 
tinued our  experiment  by  increasing  the  load  in  the  scale 
pan  still  farther  at  the  same  rate,  we  should  soon  .have 
reached  a  stress  at  which  neither  of  these  statements  are 
true.  In  other  words,  we  should  have  found,  that  while 
for  stresses  below  a  certain  limiting  value,  steel  is  elastic 
and  stretches  in  proportion  to  load,  for  stresses  exceeding 


242 


ELEMENTARY  PRACTICAL  MECHANICS 


Breaking  Load  66,200  Ibs.pera 
Elastic  Limit  37,000  Ibs.  «•  *< 
Mod.Elast.  31,200,000  «  •• 


this  limit  stretch  is  more  rapid  and  the  material  is  no  longer 
perfectly  elastic  but  takes  a  vermanent  "  set  "  i.e.,  remains 
permanently  lengthened. 

The  curves  of  Fig.  172,  which  were  obtained  from  tests 
of  an  annealed  steel  wire,  will  make  these  statements 
clearer.  This  wire  was  0.048  in.  diameter  and  86  ins.  long. 
Stretching  loads  were  applied  in  the  same  manner  as  before 
but  were  steadily  increased  until  the  wire  broke  at  109  Ibs. 
actual  pull,  after  a  total  elongation  of  about  10  ins.  The 

dotted  curves  shows  the 
results  for  loads  below  85 
Ibs.,  with  loads  plotted  to 
twenty  times  the  scale. 

This  curve  is  a  straight 
line  up  to  a  pull  of  about 
45  Ibs.,  or  27,500  Ibs.  per 
sq.in.,  and  therefore 
10'  Hooke's  Law  applies  for 
this  range.  As  the  load 
increases,  however,  the 
curve  deviates  slightly 

from  a  straight  line,  and  at  about  60  Ibs.  pull,  or  37,000  Ibs. 
per  sq.in.,  elongation  occurs  at  a  perceptibly  faster  rate 
than  increase  of  load.  This  limiting  stress  measures  the 
elastic  strength  of  the  material.  It  also  fixes  what  is  known 
as  the  elastic  limit  of  the  wire  or  the  point  at  which  the 
wire  ceases  to  be  perfectly  elastic  and  begins  to  take  set. 
As  will  be  seen  from  the  curve,  the  elastic  limit  is  approached 
gradually  and  its  exact  value  is  often  difficult  to  determine 
with  certainty. 

At  a  load  slightly  greater  than  that  at  the  elastic  limit 
(here  about  75  Ibs.  pull),  the  wire  begins  to  stretch  very 
rapidly.  The  stress  at  which  this  occurs  for  any  material 
is  known  as  the  yield  point. 


2  4  6 

.Total  Stretch  jn  Inches 


FIG.  172. — Strain  diagram  for  an  an- 
nealed steel  wire. 


ELEMENTARY  PRACTICAL  MECHANICS  243 

The  full  curve,  or  as  it  is  commonly  termed,  the  "  strain 
diagram  "  of  Fig.  172,  therefore  furnishes  a  complete  his- 
tory of  the  behavior  of  the  specimen  of  annealed  steel  wire 
when  subjected  to  steadily  increasing  tension.  It  may  be 
taken  as  in  many  ways  typical  of  the  behavior  of  all  elastic 
material  under  an  increasing  stress  of  any  character.  A 
portion  of  the  stress-strain  curve  is  a  straight  line,  which 
indicates  the  range  of  stress  under  which  the  material  is 
elastic.  The  elastic  limit  and  yield  point  of  the  material 
are  more  or  less  clearly  marked,  and  the  rapid  strain  under 
higher  stress  and  the  total  strain  when  rupture  occurs  are 
clearly  indicated.  As  the  safe  load  for  any  particular 
member  of  a  structure  must  always  be  less  than  the  load  at 
the  elastic  limit,  the  importance  to  the  engineer  of  a  knowl- 
edge of  the  behavior  of  each  kind  of  material  used  under 
the  stress  to  which  it  is  to  be  subjected  is  evident.  Ample 
allowance  must  also  be  made  for  accidental  overloads. 

breaking  load  „ 

The  ratio  — — - — -  for  any  part  of  a  structure  is 

safe  load 

called  its  factor  of  safety.  Values  for  the  factors  of  safety 
in  common  use,  and  also  for  the  ultimate  strength,  modu- 
lus of  elasticity,  etc.,  of  a  few  common  materials  will  be 
found  in  the  Appendix.  For  more  complete  data  the  stu- 
dent should  consult  any  good  engineering  handbook  or 
text  on  Materials  of  Construction. 

130.  Ultimate  Strength. — The  ultimate  strength  of  any 
material  for  tension,  compression,  or  shear,  is  the  number 
of  pounds  per  square  inch  of  tensile,  compressive,  or  shear- 
ing force  required  for  rupture.  Thus  if  a  wrought-iron  bar 
£  ins.  diameter  breaks  at  an  actual  pull  of  28,700  Ibs.,  its 
ultimate  strength  for  tension  is 


244 


ELEMENTARY  PRACTICAL  MECHANICS 


The  ultimate  strength  of  the  ordinary  materials  of  con- 
struction for  simple  tensile  or  compressive  stress  has  been 
pretty  definitely  determined.  Ultimate  strengths  for  shear, 
and  especially  for  cases  in  which  the  material  is  subjected 
simultaneously  or  in  alternation  to  more  than  one  type  of 
stress  are  not  so  definitely  known.  Safe  loads  for  such 
conditions  are  fixed  largely  by  practical  experience. 

131.  Conditions  Affecting  Strength  and  Elasticity. — The 
ultimate  strength  and  elastic  properties  of  a  specimen  de- 
pend not  only  upon  the  material  but  also  to  some  extent 


60000 


A  50000 


40000 


30000 


.40 


.10  Inches  .20  .30 

Actual  Elongations  in  Eight  Inches 

FIG.  173. — Curves  showing  elevation  of  elastic  limit  due  to  repeated 

strain. 

upon  the  previous  treatment  to  which  the  material  has 
been  subjected,  upon  its  temperature,  etc.  Repeated 
hardening  and  annealing  seems  to  strengthen  steel.  Cold- 
rolled  pieces  which  are  finished  cold  between  polished 
rollers  and  thus  reduced  in  size  (strained)  show, -on  test, 
greater  ultimate  strength  and  a  higher  elastic  limit  than 


ELEMENTARY  PRACTICAL  MECHANICS 


245 


the  original  material.  Specimens  taken  from  the  interior 
of  large,  solid  shafts  giv.e  usually  much  lower  values  for 
strength  and  elastic  limit  than  those  obtained  with  speci- 
mens taken  nearer  the  circumference,  thus  showing  the 
effect  of  difference  in  working  and  tempering. 

In  general,  any  material  loaded  repeatedly  beyond  its 
elastic  limit  with  periods  of  rest  between,  shows  on  test 
after  each  such  treatment,  that  its  elastic  limit  has  been 
raised.  Fig.  173  shows  the  results  of  such  a  series  of  tests 
performed  in  the  Strength  of  Materials  Laboratories  of  Pratt 
Institute  upon  a  Bessemer  steel  bar  18  ins.  long  and  £  in. 
diameter.  The  loads  were  applied  by  an  Olsen  testing 
machine,  and  elongations  taken  with  an  extensometer. 
After  each  test  in  which  the  stress  was  increased  beyond 
that  at  the  elastic  limit,  the  specimen  was  allowed  to  rest 
for  two  days  before  retesting,  except  for  the  fifth  and  last 
test,  which  was  continued  until  the  specimen  broke.  Fig. 
173  shows  the  results  of  the  first  four  tests.  The  elastic 
limit  was  found  to  be  higher  in  each  test  except  the  second. 

In  Fig.  174  the  fifth  test  and  a  test  of  a  second  fresh 
specimen  cut  from  the 
same  steel  bar  are  shown 
together.  The  repeated 
stresses  beyond  the  elas- 
tic limit  had  increased 
the  ultimate  strength  of 
the  material  about  15% 
and  had  raised  the 
elastic  limit,  but  the 
ductility  of  the  ma- 
terial was  decreased 
until  the  elongation  wras  nearly  34%  less. 

It  is  an  interesting  question  as  to  how  long  repeated 
straining  would  continue  to  have  the  effects  here  indicated, 


320CO 


.4    .6     .8    1.0  1.2  1.1   1.6  1.8  2.0  2.1  3.2 

Extension  in  Inches 


FIG.  174. 


246 


ELEMENTARY  PRACTICAL  MECHANICS 


and  upon  this  point,  owing  to  the  nature  of  the  problem, 
reliable  data  is  not  plentiful.*     It  is  apparent,  however,  that 

though  at  first  the  ma- 
terial is  rendered  stronger 
it  is  also  made  more 
brittle,  so  that  ultimately 
it  breaks  with  little  or  no 
elongation.  We  may  as- 
sume, therefore,  that 
every  time  a  specimen  is 
overloaded  and  given  a 
"set  "  it  is  brought  just 
so  much  nearer  its  ulti- 
mate failure. 

Figs.  175  and  176  show 
the"  results  of  some  tests, 


ANNEALED  IRON  WIRE  # 

Tensile  Strength  in  Steam  63000  Per  sq. 

<«  t.       »  Boom  Temp.06000  "    " 

««  •«       ««  Ice  Water    67000    "    " 

..       «  Liquid  air   127000  "    " 


.40          ,  .80 

Total  Elongation  of  12.08  inches 


FIG.  175. — Curves  showing  the  effect 
of  temperature  on  elastic  strength 
of  annealed  iron  wire. 


conducted  in  the  Physics 
Laboratories  of  Pratt  In- 
stitute, of  the  relative  ultimate  strength  of  iron  and  copper 
at  room  temperature  (70° F.), at  steam  temperature  (212°F.); 
in  ice  water  (about  40° 
F.),  and  in  liquid  air 
(-312°  F.).  These  J 
curves  show  that  both  « 
ultimate  strength  and  1 
elastic  limit  are  greatly 
affected  by  temperature. 


COPPER  .,,..c- 

Tensile  Strength  in  Steam       35300  fr  -ft  Per  sq.  in, 

«        "  Room  Temp.  40000 $  «     "    " 

Ice  Water      40500^  "    "    " 


JJL 


"  T.iquid  Air    60000 


NTT 


.40      .CO      .80     1.00     1.20    1.40    1.60   1.80 
Total  Elongation  of  12.08  inches         % 

The   tensile   strength  of  FIG.  176. 

iron   was    found   to    be 

about  95%  greater  at  -312°  F.  than  at  70°  F.,  and  that 

of  copper  about  50%  greater. 

*  Interesting  data  of  actual  tests  may  be  found  in  pamphlets  pub- 
lished by  the  Bethlehem  Steel  Co.,  and  in  results  obtained  by  Govern- 
ment engineers  at  the  Watertown  Arsenal. 


ELEMENTARY  PRACTICAL  MECHANICS 


247 


These  results  are  shown  even  more  strikingly  in  the 
curves  of  Fig.  177,  in  which  tensile  strength  is  plotted 
against  absolute  temperatures  Fahrenheit.  From  these 
curves  it  appears  that  the  tensile  strength  of  both  iron 
and  copper  tends  to  approach  a  minimum  (which  for  iron 
is  about  the  tempera- 
ture of  boiling  water), 
after  which  it  again 
increases.  These  re- 
sults are  borne  out 
by  tests  made  at  the 
Watertown  Arsenal,  in 
which  it  was  found 
that  for  wrought  iron 
a  minimum  tensile 
strength  is  reached  at 
a  temperature  between 
200°  F.  and  300°  F. 
(660°-760°  absolute), 
after  which  the  strength 
again  increases  as  the 
temperature  rises  until 
about  600°  F.  when  the 
strength  is  20%  to  25% 
above  normal.  From 
this  temperature  the 

tensile    strength    decreases   rapidly   with  rise  in  tempera- 
ture. 

132.  Simple  Shear. — Suppose  a  rectangular  block  of 
elastic  material,  as  ABCDEFG,  Fig.  178,  to  be  attached  to 
a  rigid  support  over  the  entire  face  EFG,  and  suppose  a 
horizontal  force  P  to  be  distributed  uniformly  over  the 
opposite  face  ABCD.  Further,  assume  that  the  force  per 
square  inch  over  ABCD  is  transmitted  uniformly,  from 


13WW 

120000 

.100000 

a 

^ 

i 
1 

i 

q   80000 

a 

If 
j 

£  60000 

I 

10000 
20000 

\ 

\ 

Anne* 

led  Ire 

nWir 

\ 

\ 

\ 

\ 

\ 

> 

^^ 

—  —  o 

~^X 

Xj 

pper  ^ 

fire 

^ 

v^ 

9 

3 

•^^ 

'  0 

I 

M 
1 

23 

r& 

200  400  600 

Temperature  in.Absolute  Degrees  Fahrenheit 

FIG.  177. 


248 


ELEMENTARY  PRACTICAL  MECHANICS 


plane  to  plane,  down  through  the  block  to  EFG.  Under 
these  assumed  conditions  the  block  will  be  under  a  simple 
shear  stress  tending  to  slide  the  particles  at  any  horizontal 
section  over  those  below,  and  the  block  will  take  the  new 
position  A'B'C'D'EFG  without  bending.*  In  other  words, 
the  block  will  undergo  a  simple  shear  strain.  If,  for  illus- 
tration, the  total  pull  P  is  100  Ibs.  and  the  area  of  surface 

ABCD  is  10  sq.ins.,  the  shear  stress  is =  10  Ibs.  per 

sq.in.     Under    the    assumed    conditions    this    is    uniform 

throughout  the  block. 

The  total  horizontal  motion  will  evidently  be  zero  at 

plane  EFG  and  will  increase  with  the  vertical  height  to  a 

maximum  at  the  plane 
ABCD.  Thus,  if  AE  = 
2ME,  the  motion  AA'  will 
equal  2XMM'.  The  hori- 
zontal movement  at  a  dis- 
tance one  inch  above  the  fixed 
plane  is  defined  as  the  shear 


FIG.  178. — Simple  shear  strain. 


strain.     Thus  if  EM  =  1  in. 

and  MM'  =0.0005  in,,  the 
shear  strain  is  0.0005  in.  per  inch.  At  a  plane  1.4  ins. 
above  EFG,  the  movement  would  be  1.4X0.0005  =  0.0007 
ins. 

In  this  case,  as  for  tension,  shear  strain  will  be  pro- 
portional to  shear  stress  until  the  elastic  limit  of  the 
material  is  reached,  and  upon  removal  of  the  stress  the 
block  will  return  to  its  rectangular  form.  Stresses  greater 

*  The  conditions  here  assumed  can  only  be  rigidly  true  when  other 
forces  are  applied  to  the  block  to  prevent  bending.  If,  however, 
the  thickness  AE  of  the  block  is  small  as  compared  to  its  length  AC, 
our  assumed  conditions  would  be  approximately  true  for  the  middle 
portion  of  the  block. 


ELEMENTARY  PRACTICAL  MECHANICS  249 

than   the   elastic  limit  will  give  a  permanent  set  to.  the 
material. 


The  ratio  -  -  —  is  called  the  modulus  of  rigidity. 
shear  strain 

The  modulus  of  rigidity,  of  steel  is  about  13,000,000. 

The  shear  stress  required  for  rupture  measures  the  ulti- 
mate strength  of  a  material  for  shear.  A  common  illustra- 
tion of  rupture  under  a  shear  stress  is  furnished  in  punch- 
ing rivet  holes  in  a  plate.  The  area  of  the  curved  surface 
of  the  hole  (circumference  of  hole  X  thickness  of  plate). 
multiplied  by  the  strength  of  the  material  of  the  plate  for 
shear  represents  the  force  with  which  the  punch  must  be 
pressed  against  the  plate. 

133.  Torsion.  —  If  we  mark  parallel  straight  lines  from 
end  to  end  upon  the  surface  of  a  flexible  rubber  rod  and 
then  clamp  one  end  fast  and  apply  to  the  other  end  a 
couple  which  tends  to  rotate  this  end  about  the  axis  of  the 
rod,  we  find  that  the  twist  is  communicated  along  the 
entire  rod  to  the  end  which  is  fast  in  such  a  manner  that 
the  straight  lines  become  helices,  about 
the  axis  of  the  rod.  The  actual  dis- 
placement of  a  point  in  any  line  from 
its  original  position  varies  directly  with 
its  distance  from  the  fixed  end,  being 
zero  at  this  end  and  a  maximum  at  the 
free  end  where  the  torque  is  applied. 
This  is  the  general  phenomenon  known  FIG.  179. 

as  twisting  or  torsion.  A  little  con- 
sideration will  make  it  clear  that  the  strain  imparted  to 
the  material  of  the  rod  is  &  shear  strain.  The  fibers  have 
been  displaced  through  an  angle  with  the  axis  of  the  rod, 
and  the  particles  at  every  section  have  been  rotated  through 
an  arc  the  center  of  which  lies  at  the  center  of  the  section. 
The  outer  particles  at  a  section  as  P,  Fig.  179,  have  moved 


250 


ELEMENTARY  PRACTICAL  MECHANICS 


to  the  new  position  Pf  a  particle  nearer  the  axis,  as  Q  has 
moved  to  the  new  position  Q',  while  particles  at  the  center 
0  have  not  been  rotated  at  all.  The  shear  strain  at  a  sec- 
tion is  greater  and  greater,  as  the  material  lies  farther  from 
the  axis  0;  it  is  also  greater  the  farther  the  section  is  taken 
from  the  fixed  end  of  the  rod. 

134.  Laws  of  Torsion. — A  simple  laboratory  apparatus 
for  the  study  of  the  laws  of  torsion  is  shown  in  Fig.  180. 
A  clamp,  which  may  be  adjusted  to  rods  of  different  diam- 


FIG.  180. — Apparatus  for  studying  laws  of  torsion. 


eter,  holds  the  end  A  of  the  rod  to  be  twisted;  the  other 
end  rests  in  a  loose  bearing  B.  Torque  is  applied  by  means 
of  cords  passed  around  the  grooved  wheel  C  and  support- 
ing the  equal  weights  W\  and  W2.  In  this  way  pressure 
upon  the  bearing  B  is  avoided.  Adjustable  mirrors,  MI, 
M2,  M%,  etc.,  are  attached  to  the  rod  by  sliding  spring 
sleeves,  at  the  equal  intervals  AM\,  MiM2l  M2M3,  etc., 
and  the  amount  of  twist  for  each  length  is  determined  by 
viewing  in  these  mirrors,  by  means  of  a  telescope,  the 
image  of  a  vertical  scale  placed  at  relatively  so  great  a 
distance  that  it  is  practically  equally  distant  from  each 
mirror.  The  twist  may  be  taken  as  proportional  to  the 


ELEMENTARY  PRACTICAL  MECHANICS  251 

reading  in  scale  divisions  for  each  mirror,  or  the  actual 
angle  of  twist  may  be  found  from  the  relation: 

Change  in  scale  reading 

-=tan.  angle  of  twist. 


Distance  scale  to  mirror  X  2 

With  this  apparatus  it  may  be  shown  experimentally 
that: 

1.  If  a  twisting  moment  T  produces  an  angle   of  twist 
<f>,  a  moment  of  2T  will  produce  a  twist  of  2<p,  etc.     Or, 
angle  of  twist  is  proportional  to  twisting  moment. 

2.  If  distance  AMi  =  MiM2  =  M2M3}  etc.,  and  twist  at 
MI  is  10  scale  divisions,  twist  at  M2  will  be  20  divisions, 
at  MZ  30  divisions,  etc.     Or,  angle  of  twist  for  the  same 
torque  is  proportional  to  length  of  rod. 

3.  For  round  rods  or  shafts  of  different  diameters,  the 
angle  of  twist  for  the  same    torque  varies    inversely   as  the 
fourth  power  of  the  diameter. 

4.  For  rods  of  the  same  size  but  different  materials,  the 

twist  for  the  same  torque  varies  as  -  —  --  -         --  .* 

modulus  of  rigidity 

*  (a)  Modulus  of  Rigidity  by  Torsion.  —  It  may  be  shown  that  the 
torque  T  required  to  twist  a  rod  of  length  I  and  diameter  D  through 
an  angle  of  <£  degrees,  is  expressed  by  the  equation  T=NIa,  where 
N  is  the  modulus  of  rigidity  for  the  material,  /  the  moment  of  inertia 
of  the  section,  a  the  twist  per  unit  of  length  expressed  in  radians,  a 

evidently  equals      -,  and  for  round  rods  /=  —  .     Therefore  N  may 


be  computed  by  the  equation, 


(Dimensions  should  be  in  inches,  T  in  pound-inches.) 

(6)  Modulus  of  Rigidity  of  a  wire  by  Torsion  Pendulum. — The  mod- 
ulus of  rigidity  of  a  wire  may  be  found  by  using  it  as  the  suspension 


252  ELEMENTARY  PRACTICAL  MECHANICS 

5.  Twisting  moments  required  for  rupture,  other  things 
being  the  same,  are  proportional  to  the  cube  of  the  diam- 
eter of  the  rods. 

PROBLEMS 

>Wr 

1.  A  cast-iron  bar  which  is  to  be  subjected  to  a  tension 
of  34,000  Ibs.  is  to  be  designed  so  that  the  unit  stress  shall 
be  2500  Ibs.  per  sq.in.     What  should  be  the  sectional  area 
in  square  inches?     If  the  bar  is  round  what  should  be  its 
diameter? 

2.  A  round  rod  of  wrought  iron  2^  inches  in  diameter, 
ruptures   under   a   tension   of   271,000   Ibs.     What   is   its 
ultimate  strength? 

3.  A  steel  eye-bar  30  ft.  long  is  1^X6  ins.  in  size.     How 
much  does  it  elongate  under  a  pull  of  90,000  Ibs.? 

for  a  heavy  disk  of  known  moment  of  inertia  about  an  axis  through 
its  center.  The  wire  should  be  attached  at  the  center  of  the  disk. 
Then  if  this  pendulum  be  set  vibrating,  without  swinging,  about  a 
vertical  axis,  and  its  period  t  determined,  modulus  of  rigidity  may 
be  found  from  the  relation  : 

Modulus  of  rigidity  of  wire  =   ,  ,    , 


where  IP  is  the  polar  moment  of  inertia  of  the  disk,  I  the  length  of 
the  wire  in  feet,  r  its  radius  in  feet,  t  the  period  of  vibration  in  seconds. 

If  the  modulus  of  rigidity  of  the  suspension  wire  is  known,  it  should 
be  noted  that  this  equation  also  enables  us  to  determine  the  moment 
of  inertia  of  bodies  which  may  be  suspended  in  place  of  the  disk 
about  an  axis  which  is  the  axis  of  the  suspension  wire  continued. 

Or,  if  the  modulus  of  rigidity  of  the  wire  is  unknown,  we  may  still 
find  the  moment  of  inertia  of  a  body  by  use  of  a  second  body  whose 
moment  of  inertia  is  known  or  readily  computed,  as  follows:  Sus- 
pend one  body  II  and  determine  the  period  ^.  Then  add  the  second, 
72,  and  determine  the  period  t2  for  the  two  combined.  Then, 


from  which  either  ^  or  72  may  be  found  if  the  other  is  known. 


ELEMENTARY  PRACTICAL  MECHANICS  253 

4.  A  bar  1  in.  square  and  2  ins.  long  elongates  0.0004  in. 
under  a  tension  of  5000  Ibs.     Compute  the  coefficient  of 
elasticity. 

5.  A  wire  1  mm.  diameter  breaks  under  pull  of  125  Ibs. 
Under  what  load  will  a  wire  of  the  same    material    and 
with  diameter  1.65  mm.  break? 

6.  A  steel  column  6  sq.ins.  in  section  and  8  ins.  long 
bears  a  load  of  75,000  Ibs.     How  much  will  it  be  com- 
pressed? 

7.  What  must  be  the  sectional  area  of  the  legs  of  the 
shears,  Fig.  91,  in   order  that,  assuming  no  bending,  the 
stress  may  not  exceed  10,000  Ibs.  per  sq.in.?     Assume  a 
load  of  100  tons,  a  20°  angle  between  legs,  a  45°  angle 
between  tie  and  horizontal,  a  15°  angle  between  plane  of 
legs  and  vertical,  and  that  the  hoisting  rope  leaves  the 
upper  block  parallel  to  the  tie. 

8.  The  tie  in  the  derrick  of  Fig.  15  has  a  tensile  strength 
of  90,000  Ibs.  per  sq.in.     It  has  3  strands  each  £  in.  in 
diameter.     What  load  at  L  will  be  required  to  break  it? 
Assume  Angle  DEC  =  100°, 

«    EDC=  40°, 
"    CDL=  40°, 
and  disregard  the  weight  of  DC. 

9.  A  plate  &  ins.  thick  has  a  shearing  strength  of  60,000 
Ibs.  per   sq.in.     What   force  will  be  required   to    drive  a 
punch  f  in.  diameter  through  the  plate? 

10.  The  tension  members  of  the  truss,  Fig.  96,  consist 
of  flat,  wrought -iron   strips,   J  in.  thick.     How  wide  must 
each  be  in  order  that  the  stress  may  not  exceed  7500  Ibs. 
per  sq.  in.?     What  will  be  the  strain  for  this  load? 

11.  A  tie  rod  28  ft.  long,  2  sq.ins.  section,  carries  a  load 
of  30,000  Ibs.     It  is  stretched  ^  in.     Find  stress,  strain, 
and  modulus  of  elasticity. 

12.  What  will  be  the  deflection  due  to  shear  in  a  beam 
8  ins.  long  and  2  sq.ins.  section,  if  it  is  fixed  at  one  end 
and  a  load  of  30,000  Ibs.  is  placed  at  the  other  end?     Assume 
a  modulus  of  rigidity  of  13,000,000. 


254 


ELEMENTARY  PRACTICAL  MECHANICS 


135.  Bending. — Fig.  181  shows  a  convenient  laboratory 
device  for  illustrating  the  general  laws  of  bending.  The 
beam  to  be  bent,  which  may  be  either  a  wooden  or  metal 
rod  of  any  desired  section,  rests  upon  adjustable  knife 
edges  supported  rigidly  upon  a  lathe  bed.  Bending  loads 
are  applied  by  scale  pan  and  weights  suspended  from  the 
middle  of  the  beam,  and  the  resulting  deflections  are  meas- 
ured by  determining  with  a  micrometer  screw  the  distance 


FIG.  181. — Apparatus  for  illustrating  laws  of  bending. 

from  a  fixed  level  to  the  surface  of  the  beam  at  the  middle. 
Where  more  precise  measurements  are  desired,  the  microm- 
eter screw  may  be  replaced  by  an  optical  lever  carried  on 
the  arm  which  here  supports  the  micrometer  screw,  the 
motion  of  which  is  measured  in  terms  of  the  divisions  of  a 
scale  placed  a  known  distance  from  the  mirror  of  the  lever. 
Results  obtained  in  the  laboratory  with  this  apparatus 
are  shown  by  Fig.  182.  It  will  be  seen  from  these  curves, 
that  the  deflection  is  proportional  to  load  when  the  elastic 
limit  of  the  beam  is  not  exceeded.  For  this  range  of  stresses 


ELEMENTARY  PRACTICAL  MECHANICS 


255 


I 

yT 

, 

' 

/ 

, 

& 

" 

/ 

/ 

S* 

j 

X 

/ 

s+ 

1 

I 

/ 

s 

^ 

1 

\ 

f 

c 

OS 

s 

A 

| 

-.J- 

4 

16 

f 

S 

Soo 

tions 

~? 

^i 

"*V5> 

H4; 

2  "' 

1 

11   1    1 

0.1           0.2          0.3           0.4          0.5        0.6 
Deflections  in  Inches 

FIG.  182. — Curves  showing  relation  of 
DEFLECTION  TO  LOAD  with  differ- 
ent WIDTHS. 


also,  the  beam  returns  to  its  original  position  when  the 
bending  load  is  removed.     If  loaded  beyond  the  elastic 
limit,  the  beam  takes  a 
permanent  "set." 

136.  Neutral  Axis.  — 
Nothing  is  shown  in  the 
preceding  experiment  as 
to  the  character  or  dis- 
tribution of  the  stress 
to  which  the  beam  is 
subjected.  If,  however, 
we  examine  a  beam 
when  bent,  we  may 
form  definite  conclu- 
sions upon  these  points. 
Fig.  183  shows  such  a 
beam  with  the  deflec- 
tion greatly  exaggerated  for  the  purposes  of  illustration. 
A  uniformly  distributed  load,  in  place  of  a  simple,  central 
load,  is  assumed  for  the  same  reason. 

It  is  apparent  from  this  diagram  that  the  edge  AB  of 
the  beam  is  sftorZened  by  the  bending,  and  that  the  edge 
CD  is  lengthened.  The  upper  portion  of  the  beam  must 

therefore  be  under 
compressive  stress,  and 
the  lower  portion 
under  tensile  stress. 
At  some  place  be- 
tween A B  and  CD, 

FIG.  183.— Diagram  of  a  loaded  beam,      as    MN,   there    must 
bending  exaggerated.  be  a  line  which,  while 

curved     because    of 

the  load,  remains  of  the  same  length  as  when  the  beam 
was    straight.      A    surface    may    therefore    be    imagined 


E(a)Gv ', 

F       H 

R2      Section 


256  ELEMENTARY  PRACTICAL  MECHANICS 

across  the  beam  through  M N,  above  which  the  fibers  are 
under  compression  and  below  which  they  are  under  ten- 
sion, but  in  which  there  is  neither  compression  nor  tension. 
This  surface  is  known  as  the  neutral  surface  for  the  beam. 
The  straight  line  XX' ',  which  shows  the  position  of  this 
surface  across  a  section  of  the  beam,  is  called  the  neutral 
axis  of  the  section. 

137.  Maximum  Tensile  and   Compressive  Stress  for  a 
Section. — The  fibers  of  the  beam  are  shortened  most  at 
AB]    therefore   at   any  section  the   compressive   stress  is 
zero  at  the  neutral  axis  XX'  and  increases  in  proportion 
to  the  distance  of  the  fiber  from  XX'  to  a  maximum  stress 
at  EG.     In  the  same  way  the  tensile  stress  on  the  other 
side  of  the  neutral  axis  is  zero  at  XX'  and  increases  with 
the  distance  to  a  maximum  at  FH.     If  a  beam  fails  through 
compression,  it  will  be  most  apt  to  yield  first  at  the  outer 
fibers  EG,  where  the  stress  is  greatest;   and  if  it  fails  be- 
cause the  tension  is  too  great,  it  will  be  most  apt  to  yield 
at   FH,  where  the  tensile  stress   is   maximum.     In  com- 
puting the  safe  load  for  a  beam,  therefore,  it  is  necessary 
only  to  consider  the  stress  at  the  outer  fibers  for  any  sec- 
tion. 

138.  Shear  in  Beams. — In  addition  to  the  tensile  and 

compressive  stresses  just  con- 
sidered, there  is  usually  a 
shear  stress  at  each  section 
of  a  loaded  beam.  The  model 
shown  in  Fig.  184  may  be 
used  to  illustrate  this.  A 
portion  CDEF  has  been  re- 

FIG.  184.-Modelofabeam.       moved    from    the   beam?   and 

to  supply  the  tensile  and  com- 
pressive forces  a  cord  has  been  inserted  at  CD  and  a  rod 
capable  of  resisting  compression  only  has  been  placed  at 


ELEMENTARY  PRACTICAL  MECHANICS  257 

EF.  If  the  end  A  is  held  rigidly,  and  a  bending  load  W 
applied  at  B,  it  is  found  that  an  additional  force,  the 
vertical  pull  W2,  must  be  introduced  at  the  section  DF 
to  support  the  outer  end  of  the  beam.  This  represents 
the  shearing  force  exertod  on  the  section  DF  by  the 
material  at  the  left.  It  may  be  shown  experimentally 
also  that  W2  =  W\  +  weight  of  end  DB  of  beam. 

The  presence  of  a  shearing  force  exerted  by  the  remain- 
ing portion  is  apparent  also  when  we  consider  the  static 
conditions  for  a  portion  of  a  beam. 
Thus  suppose  we  represent  the 
portion  of  the  beam  in  Fig.  183 
to  the  left  of  the  section  EF  as 
a  "  f  ree  body";  the  forces  act- 
ing (see  Fig.  185)  are  the  reaction 
R!  of  the  support,  the  loads  Wl9  FlG  185._Force  on  a  part 
W2,  Wz,  and  the  weight  of  the  Of  a  loaded  beam. 

beam,  and  the  molecular  forces  at 

the  section  coming  from  the  material  at  the  right.  These 
will  consist  of  a  push  S  distributed  over  a  portion  of  the 
section  and  a  pull  T  distributed  over  the  remainder. 
The  conditions  for  equilibrium  are: 


(1) 

(2)  "     F  =  0; 

(3)  "     (moments)  =0. 

We  know,  from  the  assumption  of  a  symmetrical  beam 
symmetrically  loaded,  Fig.  183,  that  Ri=R2,  and  therefore, 
that  R\  is  less  than  the  sum  of  TFi,  W2,  TF3,  and  the  weight 
of  portion  AE  of  the  beam.  To  satisfy  condition  (2), 
therefore,  the  molecular  forces  at  the  section  EF  must  have 
an  upward  j  vertical  resultant  Q  of  such  value  that: 


=  Wi  +  W2  +  W3  +  weight  of  beam  AE- 


258  ELEMENTARY  PRACTICAL  MECHANICS 

Q  therefore  represents  the  shearing  force  at  the  section  EF. 
It  is  evident  from  a  little  consideration  that  Q  is  greatest 
at  each  end  of  the  beam  where  it  is  equal  to  ^  (total  load  + 
weight  of  beam),  and  zero  at  the  middle,  since  Ri=R2  = 
sum  of  load  arid  weight  of  beam  either  side  of  the  section. 

The  student  should  assume  actual  loads  and  dimensions 
and  compute  the  value  of  the  shearing  force  at  several 
sections  along  a  beam  loaded  as  above,  and  also  for  other 
common  types  of  support  and  loading  shown  in  Fig.  186. 

139.  Bending  Moment.  —  The  sum  of  the  moments  of  all 
forces  acting  upon  the  portion  of  a  beam  between  an  end 
and  any  selected  section,  about  the  neutral  axis  of  the 
section,  is  called  the  bending  moment  for  that  section.  Thus 
referring  to  Fig.  185,  the  bending  moment  M  for  the 
section  at  EF  is  expressed  by  the  equation: 


weight  of  A  EFC  X^AE- 

The  maximum  bending  moment  for  the  beam  of  Fig.  183 
is  evidently  for  a  section  at  the  middle,  where  if  I  is  the 
length  of  the  beam, 

M  =  RI  X  \l  —  %  (total  load  +  weight  of  beam)  X  \l. 
Or,  since          RI  =  R2  =  J  (total  load  +  weight  of  beam), 
M  =  J  (load  +  weight  of  beam)L 

The  student  should  assume  actual  loads  and  dimensions 
and  compute  the  bending  moments  at  several  sections 
along  a  beam  with  loads  and  supports  as  in  Fig.  183,  also 
for  the  common  types  shown  in  the  diagrams,  Fig.  186. 
The  relative  bending  moments  of  different  sections  of  a 
beam  may  be  shown  most  conveniently  by  diagrams,  as  in 


ELEMENTARY  PRACTICAL  MECHANICS 


259 


Fig.  186,  in  which  the  bending  moments  at  successive  sec- 
tions are  represented  by  the  vertical  distances  to  the  dotted 
line. 


I.  Beam  fixed  at  one  end,  load  W  at  other. 
Maximum  bending  moment  at  A  =  Wl. 
Shearing  force  is  equal  and  opposite 
to  W  everywhere. 


II.  Beam  as  in  I,  load  W  uniformly  distributed. 
Maximum  bending  moment  at  A  =  W  X  \l. 
Shearing  force  is  maximum  at  A,  less 
toward  B. 


III.  Beam  supported  at  both  ends,  load  W  in 
middle.  Maximum  bending  moment  at 
middle,  =  JTFXiZ.  Shearing  force  is  JT7, 
A  to  middle,  -  JTF,  B  to  middle. 


IV.  Beam  as  in  III,  load  W  uniformly  distributed. 
Maximum  bending  moment  at  middle  = 
JTFZ.  Shearing  force  is  $W  at  A,  -\W 
at  J5,  zero  at  middle. 


w 


III 

W 


\E 


IV 


FIG.  186. — Diagrams  of  bending  moment. 


140.  Neutral  Axis  always  Through  the  Center  of  Gravity 
of  the  Section. — It  is  apparent  from  Fig.  185  that  all  forces 
acting  upon  the  "  free  body "  are  vertical  except  the 
molecular  forces  at  the  section  EF.  To  satisfy  the  con- 


260  ELEMENTARY  PRACTICAL  MECHANICS 

dition,  Sum  X  =  0,  therefore,  the  resultant  of  the  horizontal 
components  of  all  these  molecular  forces  must  be  zero. 

The  compressive  stress  at  any  point  is  proportional  to 
the  distance  of  the  point  from  the  neutral  axis:  we  may 
therefore  represent  it  by  Ky,  where  K  is  some  constant  for 
the  beam  and  y  is  the  distance  to  neutral  axis.  If  we 
imagine  the  total  area  under  compression  to  be  made  up 
of  an  infinite  number  of  areas  a,  so  small  that  the  com- 
pression over  each  is  uniform,  the  force  on  each  area  will 
be  Kya.  Therefore  the  total  compressive  force  for  the 
section  (which  will  be  equal  to  the  sum  of  all  these  sepa- 
rate forces)  will  be  Sum  Kya. 

In  the  same  manner  the  total  tensile  force  for  the  other 
side  of  the  neutral  axis  will  be  Sum  Ky'a,  where  y'  is  the 
distance  to  each  area  measured  in  the  other  direction  from 
the  neutral  axis. 

In  order,  therefore,  that  the  condition  Sum  X  =  0 
shall  hold  for  the  section,  we  must  have  Sum  Kya  —  Sum 
Ky'a  =  Q,  or  since  K  has  the  same  value  in  each, 


Sum  7/a  =  Sum  y'a. 

This  can  be  true  only  when  y  and  y'  are  measured  from 
an  axis  through  the  center  of  gravity  of  the  section.  (See 
Article  37.) 

This  is  a  reason  why  it  is  important  that  we  should  be 
able  to  find  the  center  of  gravity  of  a  surface.  By  finding 
the  center  of  gravity  of  the  section  of  the  beam  we  deter- 
mine the  position  of  the  neutral  axis. 

141.  General  Equation  for  Stress.  —  The  application  of 
the  third  condition  of  equilibrium,  Sum  (moments)  =0, 
to  the  free  body  of  Fig.  185  leads  us  directly  to  a  funda- 
mental equation  for  the  stress  in  beams.  By  this  condi- 
tion, referring  to  the  figure,  we  see  that  the  combined 


ELEMENTARY  PRACTICAL  MECHANICS  261 

moments  of  the  loads  Wi,  W2,  and  TF3,  the  reaction  R\ 
of  the  support,  and  the  molecular  forces  at  the  section, 
about  the  neutral  axis  of  the  section,  must  be  zero.  In 
other  words,  the  molecular  forces  must  supply  a  moment 
about  the  neutral  axis  which  shall  balance  any  difference  in 
the  moments  of  W\,  W2,  W%,  weight  of  beam,  and  RI  about 
the  same  axis. 

But  by  Article  139,  WlxPiE  +  W2XP2E  +  W3XP3#  + 
weight  of  portion  AE  of  beamXi-AE'— RiXAE  =  bending 
moment  M  for  the  section  AE.  Therefore  the  moments  of 
the  molecular  forces  at  a  section  about  the  neutral  axis  equals 
the  bending  moment  for  the  section. 

By  the   preceding    article    the   force  on  any  area  a  is 

Kya.     The  moment  of  this  force  about  the  neutral  axis  is 

,  forceXarm,  or  KyaXy  =  Ky2a.     The  total  moment  of  all 

the  forces  is  the  sum  of  the  moments  of  the  separate  forces 

or  Sum  Ky2a.     Hence  Sum  Ky'2a  =  M. 

But  by  definition  (Article  88)  Sum  (y2a)  for  any  area  = 
moment  of  inertia  /  for  that  area  about  the  axis  from 
which  y  is  measured. 

M 
Therefore,  KI  =  M,  and  K=—.    Substituting  this  value 

for  K  in  our  expression  for  stress,  or  Ky,  we  have  the 
general  equation: 

M 
Stress =—Xy, 

or,  The  stress  at  any  fiber  for  any  given  section  of  a  beam 

bending  moment 

equals    — : — X  distance    of   fiber  from 

moment  of  inertia  of  section 

neutral  axis. 

This  may  be  regarded  as  the  fundamental  equation  for 
beams.  Naturally  the  computation  would  be  made  only 


262  ELEMENTARY  PRACTICAL  MECHANICS 

for  the  maximum  stress  anywhere  in  the  beam.  This  will 
be  at  the  outer  fiber  of  the  section  at  which  the  bending 
moment  is  maximum.  Data  for  the  moments  of  inertia  of 
various  sections  may  be  found  in  engineering  handbooks. 

Example  1.  —  A  beam  10  ft.  long,  4X10"  in  section, 
resting  on  edge,  is  supported  at  the  ends  and  bears  a  dis- 
tributed load  of  200  Ibs.  per  foot.  What  will  be  the  maxi- 
mum stress  and  the  shearing  force? 

Maximum  bending  moment  is  at  the  middle,  and  is 

\wl  =  j-  X  200  X  10  X  120  =  30,000  in.-lbs. 

M3_4X10X10X10_1000 
T2~  12  ~~3~' 

Therefore, 

M          30,000 

Maximum  stress  =—  X5  =  —  —  -  X5=450  Ibs.  per  sq.m. 
L  luuU 

3 
Maximum  shearing  force  is  at  the  ends,  and  is 


Example  2.  —  A  round  steel  rod  18  ins.  long  in  the  clear, 
and  1J  ins.  diameter,  is  fixed  at  one  end.  What  load  may 
be  suspended  from  the  other  end  if  the  maximum  allow- 
able stress  is  5000  Ibs.  per  sq.in.? 

Maximum  bending  moment  is  at  end  which  is  fixed  and 
is  T7X18  in.-lbs. 

891 


64     3584 


.ELEMENTARY   PRACTICAL  MECHANICS  263 

Therefore, 

TFX18 
5000=  -  Xf,  from  which  TF=92.1  lbs.x 


3584 

142.  Modulus  of  Elasticity  by  Bending.  —  The  modulus  of 
elasticity  of  a  material  may  be  determined  experimentally 
from  measurements  of  the  deflection  of  a  strip  or  rod  under 
a  load  which  gives  a  tensile  and  compressive  stress  less 
than  that  at  the  elastic  limit.  If  W  is  the  load,  I  the  length 
of  the  beam  in  inches,  d  the  deflection  in  inches  due  to  W, 
I  the  moment  of  inertia  of  the  section,  it  may  be  shown 

wP 

that  modulus    of    elasticity   E  =  —-  for  a  beam  fixed  at 

oal 

one  end,  with  load  w  at  the  other. 

For  a  beam  supported  at  both  ends  with  load  w  at  the 
middle,  the  corresponding  equation  is: 


4Stf 

143.  Effect  of  Dimensions  of  a  Beam  upon  its  Stiffness. 

— It  may  be  shown  that  with  other  conditions  constant, 
the  resistance  of  a  beam  to  bending  varies  directly  with 
the  breadth,  inversely  with  the  cube  of  the  length,  and 
directly  as  the  cube  of  the  depth;  or 

bd* 
Stiffness  oc  — -. 

i 

An  experimental  study  of  these  laws,  with  the  apparatus 
shown  in  Fig.  181,  forms  an  instructive  laboratory  exer- 
cise. The  beams  should  be  straight  grained  wooden  rods 
or  metal  strips  of  rectangular  section,  carefully  dressed  to 


264 


ELEMENTARY  PRACTICAL  MECHANICS 


size,  and  the  loads  used  should  be  well  below  those  at  the 
elastic  limit.     Fig.  182  shows  the  results  of  a  test  of  the 

effect  of  increasing  the 
breadth  of  the  beam. 
Two  rods  of  the  same 
material,  and  of  the 
same  length  and  depth, 
were  used,  the  breadth 
of  one  being  twice  that 
of  the  other.  The 
breadths  were  therefore 
in  the  ratio  J.  The  de- 
flections for  the  same 
load  were:  For  10  ozs., 

.175   i  .363 

-,  for  20   ozs., 


FIG.  187. — Curves  showing  relation  of 
deflection  to  depth.  Material  of 
beam-pine,  length  40  inches. 


.090'  '  .187 

These  are  very  nearly  in  the  ratio  f,  showing  that 
the  beam  of  twice  the  width  was  bent  only  J  as  much, 
or  in  other  words,  was  twice  as  stiff. 

Fig.  187  shows  results 
of  a  similar  test  for  the 
effect  of  depth  on  stiff- 
ness. The  depths  were 
in  the  ratio  f ;  the  de- 
flections for  the  same 
load  at  the  middle  of  the 
beam  were  for  20  ozs., 

.025  .050 

— ,  for  40  ozs.,  --. 

These    ratios    are    both 

I         FIG.  188. — Curve   showing  relation  of 
approximately  |  or  — - ,         deflection  to   length.      Material   of 

W  beam-pine, 

showing      that     if     the 
depth  is  doubled  the  beam  is  (2)3  or  8  times  as  stiff. 


40 
J30 

.a 
f20 

10 
0 

==" 

e- 

. 

^-^ 

—  — 

— 

.g, 

^• 

-^ 

/ 

of 

/ 

/ 

/ 

CRC 

)SS 

/ 

ECTION 

e« 

«3<fr* 

1 

1 

0.1  0.2 

Deflections  in  Inches 


ELEMENTARY  PRACTICAL   MECHANICS  265 

Tests  of  deflections  of  different  lengths  of  the  same 
beam  with  the  same  load  are.  shown  in  the  curve  of  Fig. 
188.  The  load  was  kept  central  and  the  supports  ad- 
justed until  successively  40,  30,  20,  and  10  ins.  apart. 
The  lengths  of  beam  were  therefore  in  the  ratios  f,  f,  and  -f ; 

.310     .119  .038 

the  corresponding  deflections  wTere  , ,  and  , 

.005     .005  .005 

.  .      64    27              8                          (4)3     (3)3 
or    approximately   — ,  — ,   and    — .     Inese   are  ,   , 

(2)3 

— ,  showing  that  bending  increased  (and  therefore  stiffness 

diminished)  in  proportion  to  the  length  cubed. 

Very  suggestive  results  may  be  obtained  also  with  beams 
other  than  rectangular  in  section.  Here  it  should  be 
remembered  that  the  sections  being  similar,  breadths  refer 
to  the  relative  lateral  dimensions  of  the  beams,  depths  to 
the  relative  vertical  dimensions. 

144.  Effect  of  Dimensions  and  Manner  of  Loading  upon 
Strength  of  a  Beam. — It  has  been  shown  that  while  the 
stiffness  of  a  beam  increases  as  the  depth  cubed,  the  strength 
increases  only  as  the  depth  squared.  The  strength  is  also 
proportional  to  the  breadth  of  the  beam,  and  inversely 
proportional  to  its  length.  Or, 

bd* 

Strength  oo— . 
L 

Since  we  may  expect  rupture  to  occur  when  the  bending 
moment  at  any  section  of  a  beam  exceeds  a  certain  limiting 
value,  the  diagrams  of  bending  moment,  Fig.  186,  show 
some  interesting  relations  of  the  maximum  load  which  a 
given  beam  will  support  to  the  manner  in  which  this  load 
is  applied.  Thus  a  beam  of  given  size  and  length  which 
will  support  a  maximum  load  w  when  fixed  at  one  end  and 


266  ELEMENTARY  PRACTICAL  MECHANICS 

loaded  as  in  diagram  I,  may  be  expected  to  carry  a  load 
2w  distributed  as  in  II,  4w  if  both  ends  are  supported  and 
the  load  concentrated  at  the  middle  of  the  beam  as  in 
diagram  III,  and  Sw  if  supported  at  both  ends  with  the 
load  uniformly  distributed  as  in  diagram  IV. 

146.  Shapes  of  Beams  and  Columns. — Short  columns,  or 
columns  prevented  from  bending,  may  be  expected  to 
yield  only  at  a  stress  equal  to  the  ultimate  compressive 
strength  of  the  material.  Long  columns  not  supported 
laterally  yield  by  bending.*  It  is  desirable  therefore  to 
shape  the  section  of  posts  and  struts  to  offer  the  greatest 
possible  resistance  to  bending.  This  is  accomplished,  as  is 
seen  from  the  laws  for  stiffness  just  stated,  when  the  material 
is  distributed  as  widely  as  possible  about  the  axis  of  the 
column. 

In  beams  the  top  must  resist  compressive  stress  and  the 
bottom  tensile  stress.  The  bottom  flanges  may  therefore 
be  thin  plates  as  a  pull  does  not  produce  bending,  but  the 


*  Strength  of  Struts.  —  The  breaking  load  F  for  a  short  strut  or  one 
prevented  from  bending  is  expressed  by  the  equation  F=fA,  where 
/  is  ultimate  compres&ive  strength  of  the  material,  A  is-  area  of  sec- 
tion in  square  inches. 

Various  formulae  are  in  use  for  struts  which  yield  by  bending,  based 
upon  the  shape  and  manner  of  holding  at  the  ends.  For  a  strut  with 
what  may  be  called  "  free  "  ends,  the  load  F,  in  line  with  the  axis 
of  the  piece  which  will  produce  bending  is, 


where  E  is  Young's  modulus,  /  the  moment  of  inertia  of  a  section 
about  the  line  through  its  center  of  gravity,  I  one-half  the  length  of 
the  strut.  For  this,  and  similar  formula  for  other  types,  the  student 
is  referred  to  advanced  treatises  in  Mechanics. 

Where  the  value  of  F,  obtained  by  such  formulae,  is  greater  than 
that  resulting  from  use  of  the  formula  .F=fA,  the  lower  value  must 
be  taken  as  the  strut-  will  yield  to  compression. 


ELEMENTARY  PRACTICAL  MECHANICS 


267 


(a) 


(b) 

FIG.  189. — Sections  of 
girder. 


top  must  be  in  the  form  of  a  hollow  tube  or  shaped  thus 
O  n  to  resist  the  tendency  to  bend  under  compres- 
sion. 

Cast  iron  has  a  compressive  strength  nearly  4J  times  as 
great  as  its  tensile  strength.  The  section  of  cast  iron  gir- 
ders is  therefore  usually  made  so 
that  the  area  under  tension  is  ap- 
proximately four  and  a  half  times 
as  large  as  that  under  compression. 
Fig.  189  (a)  shows  some  common 
sections  for  cast-iron  girders. 

Steel  and  wrought  iron  have 
practically  the  same  ultimate 
strength  for  both  tension  and  com- 
pression. Symmetrical  and  econom- 
ical "  built-up "  sections  similar  to 
those  of  Fig.  189(6)  are  therefore 
common. 

In  general  the  function  of  the  flanges  is  to  resist  bending 
moment,  that  of  the  web  to  resist  the  shearing  force.  Any 
web  strong  enough  to  hold  the  flanges  in  position  and  to 
permit  the  girder  to  be  handled  is  usually  amply  sufficient 
to  withstand  the  shearing  force. 

In  beams  of  uniform  section,  only  the  maximum  bend- 
ing moment  need  be  considered.  The  section  must  be  de- 
signed to  safely  sustain  this.  Girders  of  "  equal  strength  " 
may  be  designed  in  which  the  section  varies  according  to 
variation  in  bending  moment. 

146.  Stress  for  Hooks,  Brackets,  etc. — In  the  case  of 
hooks,  brackets,  etc.,  the  applied  load  acts  in  a  line  which 
does  not  coincide  with  the  axis  of  the  piece.  The  material 
is  therefore  subjected  to  both  a  tensile  (or  compressive) 
force  and  a  bending  moment.  For  example,  in  a  hook,  as 
Fig.  190,  we  may  consider  the  stress  at  the  section  AB  as 


268 


ELEMENTARY  PRACTICAL  MECHANICS 


due  to  the  bending  moment  WxCG,  where  G  is  an  axis 
perpendicular  to  the  plane  of  the  paper 
through  the  center  of  gravity  of  the  sec- 
tion, plus  a  tensile  stress,  spread  uni- 
formly over  the  section,  equal  to 

W 

• .     The  stress  at  any  fiber 

area  of  section 

in    the    section    is     computed    as    for 
beams. 

If  the  distance  CG  is  large,  as  in  the 
case  of  many  brackets,  stress  due  to 
bending  moment  alone  is  important. 


FIG.  190. 


PROBLEMS 

1.  A  beam  AB,  16  ft.  long,  weighing  200  Ibs.  per  ft.,  is 
supported  at  both  ends.     It  bears  loads  as  follows:  5  tons 
4  ft.  from  end  A,  10  tons  at  middle,  6  tons  5  ft.  from  end 
B.     Compute  bending  moments  due  to  loads  and  weight 
of  beam  combined  for  sections  2  ft.,  4  ft.,  8  ft.,  and  10  ft. 
from  A. 

2.  In  a  beam  as  diagram  II,  Fig.  186,  beam  is  6  ft.  long 
and  load  500  Ibs.  per  foot.     Compute  maximum  bending 
moment  and  shearing  force;  also  bending  moments  2  and 
4  ft.  from  fixed  end. 

3.  The  top   and  bottom   flanges   of  a   girder  are   each 
6Xf".     Web  is  £"  thick  and   depth    of   beam  over  all  is 
10  ins.     What  bending  moment  will  be  required  to  give  a 
maximum  tensile  stress  of  10,000  Ibs.  per  sq.in.? 

4.  What  is  the  greatest  stress  in  a  bar  f  in.  diameter, 
6  ins.  long,  when  it  is  fixed  at  one  end  and  50  Ibs.  are  sus- 
pended from  the  other  end? 

5.  What  will  be  the  greatest  stress  in  a  bar  2  sq.ins.  in 
sectional  area  for  a  bending  moment  of  20,000  in.-lbs.,  if 
(a)  round?  (b)  square? 


ELEMENTARY  PRACTICAL  MECHANICS  269 

6.  A  rectangular  bar  5  ft.  long  is  supported  at  the  ends. 
It  is  2  ins.  wide  and  4  ins.  deep.     What  load  at  the  middle 
will  be  required  to  give  a  maximum  tensile  stress  of  10,000 
Ibs.  per  sq.in.? 

7.  A  beam  20  ins.  long,  2  ins.  deep,  and  1  in.  wide  is 
planed  down  so  that  its  depth  is  1J  ins.  and  breadth  £  in. 
How  is  its  stiffness  affected?     Increased  or  decreased,  and 
by  how  much? 

If  it  were  made  5  ins.  shorter  what  would  be  its  stiffness 
in  terms  of  original  stiffness? 

8.  In  a  hook,  as  Fig.  190,  the  section  at  AB  is  circular 
|  in.  diameter.     CG  is  1  in.,  W  is  400  Ibs.     What  will  be 
the  maximum  stress  at  the  section? 


CHAPTER  XIII 
MECHANICS  OF  FLUIDS 

147.  Fluids. — Matter  may  exist  in  the  form  of  solids,  as 
iron,  wood,  ice,  etc.,  or  in  the  form  of  fluids,  as  water,  oil, 
steam,  oxygen,  air,  etc.  Fluids  may  be  either  liquids  or 
gases. 

The  distinguishing  property  of  a  solid  is  that  it  has  a 
definite  shape  and  volume  which  cannot  be  changed  with- 
out the  application  of  an  external  force.  Thus  a  piece  of 
iron  has  a  particular  shape  and  an  unchanging  volume 
wherever  it  is  placed.  A  fluid,  on  the  other  hand,  has  no 
definite  shape,  but  moulds  itself  to  the  shape  of  the  vessel 
that  contains  it.  Thus  we  may  have  a  given  quantity  of 
water  or  of  air  in  a  vessel  and  may  transfer  it  to  another 
vessel  to  whose  shape  it  will  conform  itself.  Also  a  solid 
cannot  be  stirred  or  cut  without  the  application  of  con- 
siderable force,  but  a  fluid  offers  very  little  resistance  to 
the  disturbing  force.  These  examples  show  that  a  fluid  is 
composed  of  particles  that  are  mobile,  i.e.,  they  move  freely 
among  each  other  and  can  easily  be  separated. 

Furthermore,  if  we  take  a  straight  cylindrical  tube  open 
at  one  end  and  closed  at  the  other,  and  fit  it  into  a  smooth 
rod  of  some  solid  substance  such  as  iron  or  wood,  and  press 
the  end  of  the  rod  in  the  direction  of  its  length,  the  pres- 
sure applied  is  transmitted  to  the  closed  end  of  the  tube 
without  producing  any  effect  on  the  curved  surface;  but 
if  the  tube  contains  a  fluid,  either  a  liquid  or  a  gas,  and 

270 


ELEMENTARY  PRACTICAL  MECHANICS  271 

pressure  be  applied  to  it  by  means  of  a  tight  piston  fitting 
into  the  tube,  pressure  will  be  felt  not  only  on  the  base  of 
the  vessel  but  also  at  all  points  of  the  curved  surface. 
Thus  fluid  bodies  are  those  that  cannot  sustain  a  longitu- 
dinal pressure,  however  small,  without  being  supported  by 
a  lateral  pressure  also. 

148.  Distinction   between  Liquids  and  Gases. — The  dif- 
ference between  a  liquid  and  a  gas  is  that  liquids  have  a 
definite  volume  but   no   definite  shape,   while   gases  have 
neither  a  definite  volume  nor  a  definite  shape.     Thus  water 
is,  like  all  true  liquids,  practically  incompressible,  but  will 
adapt  itself  to  the  shape  of  whatever  vessel  it  is  put  into. 
But  gas,  like  air  or  illuminating  gas,  has  no  definite  shape 
and  may  also  be  expanded  by  the  lessening  of  the  pressure 
upon  it  or  compressed  to  any  extent.     These  properties  of 
solids,  liquids,  and  gases  may  also  be  expressed  as  follows: 

SOLIDS  have  elasticity  of  both  shape  and  volume. 
LIQUIDS  have  no  elasticity  of  shape,  but  perfect  elas- 
ticity of  volume. 

GASES  have  elasticity  of  neither  shape  nor  volume. 

It  should  be  noticed  that  the  distinctions  above  are  of  a 
general  nature  only;  there  are  some  substances  that  exist 
in  an  intermediate  state,  like  coal  tar,  thick  glue,  etc. 
Also  there  are  vapors  which  can  be  made  to  pass  by  im- 
perceptible degrees  from  the  gaseous  to  the  liquid  state. 
There  are  no  strict  lines  of  division  between  the  three 
states  of  matter.  The  same  substance  may,  under  differ- 
ing conditions  of  temperature,  pressure,  etc.,  exist  in  any 
one  of  the  three  states,  as  steam,  water,  and  ice.  Iron 
which  is,  as  we  know  it,  either  a  solid  or  a  liquid,  exists  in 
the  sun  only  as  a  gas. 

149.  Density. — Iron  is  more  dense  than  wood  and  water 
is  more  dense  than  air.     By  these  statements  we  mean  that 


272  ELEMENTARY  PRACTICAL  MECHANICS 

a  given  volume  of  one  substance  has  more  weight  (or  mass) 
than  the  same  volume  of  the  other. 

When  we  wish  to  state  the  density  of  a  substance  we  give 
the  weight  of  a  unit  volume  of  it.  Thus  the  density  of  iron 
is  7.8  gm.  per  cubic  centimeter;  the  density  of  water  is 
62.5  Ibs.  per  cubic  foot,  and  of  glass  is  1.5  oz.  per  cubic  inch. 
Expressed  algebraically  this  definition  is  as  follows: 

Weight 

Density  =  — —  — :  therefore  Weight  =  Volume  X  Density. 
Volume 

A  knowledge  of  the  density  of  the  common  materials  of 
construction  is  very  valuable  to  designers  and  builders  be- 
cause it  enables  them  to  compute  the  approximate  weight 
of  any  particular  part  of  a  machine  or  structure.  The 
volume  of  a  part  determined  from  the  scale  drawingX 
the  density = weight  of  finished  part.  The  densities  of  a 
few  of  the  more  common  materials  of  construction  are 
given  in  the  Appendix. 

150.  Specific  Gravity. — By  the  SPECIFIC  GRAVITY  of  a  sub- 
stance we  mean  its  relative  density  as  compared  to  water; 
or  in  other  words,  how  many  times  heavier  a  particular  body 
is  than  an  equal  volume  of  water.*  Thus  we  say  the  specific 
gravity  of  brass  is  8.4;  meaning  that  brass  is  8.4  times  as 
heavy  as  water.  A  cubic  foot  of  brass,  therefore,  weighs 
approximately  8.4X62.5  =  525  Ibs. 

Weight  of  the  substance 
bpecmc  gravitv  =  - 


weight  of  an  equal  volume  of  water 


*  Strictly,  the  comparison  should  be  to  an  equal  volume  of  pure 
water  at  4°  C.,  the  temperature  at  which  water  is  most  dense.  The 
density  of  water  changes  so  slightly  with  change  of  temperature,  how- 
ever, that  for  practical  purposes  the  comparison  may  be  made  at  any 
convenient  temperature. 


ELEMENTARY  PRACTICAL  MECHANICS  273 

If  we  use  the  metric  units,  the  density  of  a  body  and  its 
specific  gravity  are  expressed  by  the  same  number.  Thus 
the  density  of  lead  is  11.35  gm.  per  cc.,  and  its  specific 
gravity  is  11.35.  But  if  we  are  using  the  English  units,  we 
should  say  that  the  specific  gravity  of  lead  is  11.35  and 
therefore  its  density  is  11.35X62.5  =  709  Ibs.  per  cubic  foot. 

PROBLEMS 

(Consult  tables  for  required  density  or  specific  gravity.) 

1.  Find  the  weight  of  a  bar  of  copper  whose  cross-section 
is  14  sq.cm.  and  length  45  cm. 

2.  Find  the  size  of  a  round  iron  shot  to  weigh  16  Ibs. 

3.  Find  the  weight  of  a  column  of  mercury  30  ins.  high 
and  having  a  cross-section  of  ^  sq.in. 

4.  What  is  the  weight  of  the  air  in  a  room  20  X 30  X 12  ft.? 

5.  The  specific  gravity  of  sea  water  is  1.03.     How  many 
cubic  feet  of  water  are  displaced  by  a  400-ton  ship? 

6.  Find  the  weight  of  a  gallon  of  sulphuric  acid,  specific 
gravity  1.84. 

7.  Four  cubic  feet  of  cork  weigh  60  Ibs.     What  is  specific 
gravity  of  the  cork? 

8.  What  will  be  the  weight  of  40  ft.  of  lead  pipe,  inside 
diameter  2  ins.,  outside  diameter  2f  ins.? 

9.  What  weight  of  sheet  copper  ^  in.  thick  will  be  re- 
quired to  line  a  rectangular  tank  5.0  ft.  long,  3.0  ft.  deep, 
2^  ft.  wide,  inside  dimensions? 

10.  Calculate  the  weight  of  a  cast-iron  pipe  10  ft.  long, 
4.0  in.  bore,  walls  ^  in.  thick,  having  a  circular  flange  at 
each  end  9.0  ins.  diameter,  &  in.  thick. 

151.  Pressure. — In  our  discussion  of  the  mechanics  of 
fluids  we  shall  use  the  term  pressure  always  in  the  sense 
of  intensity  of  pressure,  or  pressure  per  unit  of  area.  The 


274 


ELEMENTARY  PRACTICAL  MECHANICS 


unit  of  area,  unless  specifically  stated  otherwise,  is  under- 
stood to  be  a  square  inch.  Thus  the  expression,  "  Air  at 
30  Ibs.  pressure,"  signifies  air  which  exerts  30  Ibs.  pressure 
on  each  square  inch  of  surface,  etc. 

152.  Pressure  in  Fluids  at  Rest. — Suppose  a  small  area, 
a,  Fig.  191,  to  be  horizontal  and  a  depth  Z  below  the  sur- 
face of  a  liquid  at  rest.  Assume  all  other  sources  of  pressure 

removed.     Then   the  area 

will  be  pressed  vertically 
downward  by  a  force  equal 
to  the  weight  of  the  column 
of  liquid  above  it.  This 
weight  will  be  equal  to  the 
volume  Za  of  liquid  times 
its  density  s  or  Zas.  If  p 
be  the  pressure  at  any  point 
of  the  area,  then  pa,  or 
the  total  force  on  the  area, 
must  equal  Zas,  or  p  =  Zs. 
FIG.  191.  Any  change  in  the  depth 

Z,    or   in   the    density    of 

the  liquid,  will  produce  a  corresponding  change  in  the 
weight  of  the  column  of  liquid  resting  upon  a  given 
area,  and  hence  upon  the  pressure.  Thus  we  have  the 
rule: 

PRESSURE  IN  A  LIQUID  VARIES  WITH  THE  DEPTH  AND 
WITH  THE  DENSITY  OF  THE  LIQUID. 

It  is  because  of  this  greater  pressure  as  the  distance  be- 
low the  surface  increases  that  dams,  retaining  walls  of 
reservoirs,  etc.,  are  made  thicker  and  stronger  at  the  bot- 
tom, as  in  Fig.  200. 

If  the  area  a  in  the  preceding  illustration  is  not  hori- 
zontal, the  pressure  at  some  parts  is  greater  than  at  others, 
but  the  pressure  at  any  given  point  in  the  area  is  as  before 


ELEMENTARY  PRACTICAL  MECHANICS 


275 


dependent  only  upon  the  depth  of  this  point  below  the 
surface  and  upon  the  density  of  the  liquid. 

153.  Pressure  Equal  in  all  Directions. — It  is  evident  that 
the  pressure  at  the  depth  Z  acts  equally  in  every  direction 
up,  down,  and  at  every  angle.    This  must  be  true,  for  other- 
wise the  liquid,  being  acted  on  by  an  unbalanced  force  in 
some  direction,  would  move,  which  is  contrary  to  the  sup- 
position that  the  liquid  is  at  rest,  i.e.,  in  equilibrium. 

154.  Direction  of  Pressure  on  the  Walls  of  the  Vessel. — 
It  may  also  be  seen  that  the  pressure  of  a  liquid  on.  the 
surface  of  the  containing  vessel  is  always  perpendicular  to 
that  surface,  for  if  it  were  not,  there  would  be  a  component 
of  the  pressure  tending  to  urge  the  liquid  along  the  sur- 
face and  the  liquid  being  free  to  move,  would  slide  along 
the  surface  which,  again,  is  contrary  to  the  supposition 
that  the  liquid  is  in  equilibrium. 

155.  Transmission  of  Applied  Pressure  through  Fluids. — 
If  we  take  a  vessel  full  of  water  (see  Fig.  192),  having  var- 


lp 


FIG.  192. 


FIG.  193. 


ious  apertures  of  the  same  size  fitted  with  water-tight 
pistons,  and  if  one  of  these  pistons  be  pressed  downward 
with  a  force  P  an  additional  pressure  equal  to  P  will  be 
required  at  each  of  the  other  pistons  to  preserve  equi- 


276  ELEMENTARY  PRACTICAL  MECHANICS 

librium.  Thus,  if  a  force  of  5  Ibs.  is  applied  to  P  it  will 
be  necessary  to  apply  a  force  of  5  Ibs.  at  each  of  the  other 
pistons  to  prevent  their  moving,  and  this  will  be  so  no 
matter  at  what  part  of  the  vessel  these  pistons  are  placed. 
Thus  we  see  that  any  increase  of  pressure  applied  at  one 
part  is  transmitted  equally  through  the  fluid.  Furthermore  if, 
as  in  Fig.  193,  the  pistons  are  of  different  areas,  that  at  B 
being  twice  as  large  area  as  that  at  A,  and  that  at  C  being 
three  times  as  large,  then  for  every  pound  applied  at  A 
2  Ibs.  must  be  added  for  equilibrium  at  B,  and  3  Ibs.  at  C. 
In  general,  the  forces  at  the  different  pistons  will  vary 
directly  as  the  areas  of  those  pistons,  or  if  the  pistons  are, 
as  usual,  of  a  circular  cross-section,  the  forces  will  vary  as 
the  squares  of  the  diameters.  Hence  we  arrive  at  the  im- 
portant principle: 

PASCAL'S  PRINCIPLE. — When  pressure  is  communicated  to 
any  part  of  a  fluid  it  is  transmitted  equally  on  equal  areas  in 
all  directions  through  the  fluid. 

In  the  illustrations  above  a  liquid  was  referred  to,  but 
the  student  should  notice  that  Pascal's  Principle  applies  to 
all  fluids,  both  gases  and  liquids. 

156.  Applications  of  Pascal's  Principle. — The  principle 
just  given  explains  the  action  of  many  common  and  im- 
portant pieces  of  apparatus  and  many  useful  machines. 

Suppose  we  have  two  communicating  vessels  containing 
water,  Fig.  194,  one  of  which  is  much  larger  than  the  other. 
The  vessels  are  fitted  with  pistons  P  and  p}  the  areas  of 
which  we  will  suppose  to  be  A  and  a.  If  now,  weights  W 
and  w  be  placed  on  these  two  pistons  respectively,  so  as  to 
counterbalance  each  other,  it  will  be  found  that 

W :  w  =  A  •  a, 
which  is  in  accordance  with  Pascal's  Principle. 


ELEMENTARY  PRACTICAL  MECHANICS 


277 


We  see,  also,  that  if  the  piston  p  be  pressed  down  through 
the  distance  S,  the  water  contained  in  the  smaller  vessel 
will  be  caused  to  pass  into  the  larger,  and  force  up  the 
piston  P  through  some  distance  Sf,  such  that 


since  the  volume  of  water  that  is  removed  from  one  vessel 
is  the  same  as  that  which  enters  the  other  vessel.     Hence 

a     S' 

—  = — ,  that   is,  the    sim- 

«•.  .  o  A      j, 

ultaneous  distances  moved 
by  the  pistons  vary  in- 
versely as  their  areas. 
Also 


1 

/      w     \ 

! 
i 

i 

'%%B^ 

a 

p 

Area  A 

that  is,  the  work  done  by 

w  =  work  done  by  W.     This  is  the  Hydraulic  Press. 

In  order  to  make  the  action  of  the  hydraulic  press  con- 
tinuous and  to  provide  the  motion  at  W  necessary  in  prac- 
tice, the  smaller  piston  is  operated  as  the  plunger  of  a  pump 
by  means  of  which  water  is  pumped  into  the  reservoir. 
Other  forms  of  hydraulic  and  pneumatic  machinery,  as 
hydraulic  jacks,  pneumatic  drills,  riveters,  air-brakes,  etc., 
act  upon  a  similar  principle  of  transmission  of  pressure. 

Among  common  phenomena  the  pumping  up  of  bicycle 
and  automobile  tires  with  a  small  hand  pump,  and  the  rise 
of  water  to  the  same  level  in  a  closed  system  of  pipes,  may 
be  mentioned  as  instances  in  which  Pascal's  principle 
applies. 

Fig.  195  shows  the  case  of  two  open  tubes  of  different 
size  communicating  by  a  flexible  diaphram  0,  one  contain- 


278 


ELEMENTARY  PRACTICAL  MECHANICS 


ing  mercury  the  other  water.     The  pressure  at  0  is  equal 
in  all  directions  (liquids  in  equilibrium),  therefore  Pi=P2. 

PI  =  atmospheric  pressure  Pa  + 
HI  X  dm.  P2  =  Pa  +  H2  X  dw,' 
where  dm  and  dw  =  density  of 
mercury  and  water  respec- 
tively. Therefore,  Pa  +  Hidm  = 
Pa  +  H2dw.  Whence,  Hidm  = 
H2dw  and 

Hi  _dw 
H2     dm 


FIG.  195. 


or  the  heights  at  which   the   two 
liquids    stand    in    the    tubes   are 
inversely  proportional  to  their  densities. 


PROBLEMS 

1.  TWQ    communicating   vessels    contain    fluid,    and   are 
fitted  with  pistons,  the  diameters  of  which  are  2  and  8  ins. 
respectively.     If  a  weight  of  3  Ibs.  is  placed  on  the  smaller 
piston,  what  weight  must  be  placed  on  the  larger  to  pre- 
serve equilibrium? 

2.  A  narrow  vertical  pipe  is  attached  to  a  vessel  which 
is  fitted  with  a  piston  the  area  of  which  is  200  sq.cm.     If 
the  vessel  and  pipe  contain  water,  find  the  height  of  the 
water  in  the  pipe  when  a  weight  of  40  kgm.  is  placed  on 
the  piston. 

3.  A  cylindrical  vessel  contains  mercury  to  the  height  of 
2.0  ins.  above  the  base  and  a  layer  of  8.0  ins.  of  water  rest- 
ing on  the  mercury.     Find  the  pressure  at  any  point  in 
the  base.     Mercury  is  13.6  times  as  heavy  as  water. 

4.  In  a  hydraulic  press,  diameter  of  piston  in  the  small 
cylinder  is  1^  ins.,  diameter  piston  of  large  cylinder  16  ins. 
If  water  is  forced  in  under  pressure  of  80  Ibs.  for  the  entire 


ELEMENTARY  PRACTICAL  MECHANICS 


279 


area  of  the  small  piston  what  will  be  (a)  pressure  per  square 
inch  on  larger?  (6)  total  pressure? 

5.  Bicycle  tire  is  cylindrical,  1^  ins.  inside  diameter,  and 
5  ft.  long.     Opening  for  pump  is  \  in.  diameter.     If  air  is 
forced  in  here  under  a  pressure  of 

one  pound  for  this  area,  what  will 
be  (a)  total  pressure  produced  on 
inside  of  tire?  (b)  Pressure  per 
square  inch?  (c)  Pressure  tending 
to  burst  tire  if  atmospheric  pressure 
is  15  Ibs.  to  square  inch? 

6.  What     is    the     pressure     per 
square  inch  due  to  water  alone  at 
the  bottom  of  a  pond  30  ft.  deep? 

7.  Pipe   A,  Fig.   196,   is    1.0  in.  FIG.  i96. 
inside  diameter.      It  is  filled  with 

water  to  a  height  #  =  4.0  ft.  What  weight  of  mercury  must 
be  placed  in  B  in  order  that,  when  valve  V  is  opened,  the 
liquids  will  remain  at  the  same  level? 

157.  Whole  Pressure  of  a  Liquid  on  the  Base  and  Sides 
of  the  Vessel  Containing  It. — The  laws  of  fluid  pressure 
stated  in  the  preceding  articles  enable  us  to  determine  the 

total  pressure  on  the  walls 
of  vessels,  reservoirs,  etc., 
whatever  their  shape. 

(a)  Whole  Pressure  on 
Base. — Three  cases  may  be 
distinguished,  according  as 

the  side  walls  are  vertical 
FIG.  197.  i          r  xu       u 

or    slope    from    the    base 

outward  or  inward.      See  Fig.   197. 

Suppose  the  base  in  each  case  to  be  horizontal.  If  the 
sides  are  vertical  the  whole  pressure  is  evidently  on  the 
base  and  the  pressure  is  equal  to  the  weight  of  the 
liquid. 


280 


ELEMENTARY  PRACTICAL  MECHANICS 


Next,  suppose  the  side  to  slope  outward.  In  this  case 
the  pressure  is  less  than  the  weight  of  the  liquid.  It  is  equal 
to  the  weight  of  the  liquid  within  the  dotted  lines.  What 
supports  the  liquid  between  the  dotted  lines  and  the  sides? 
In  the  third  case  where  the  sides  slope  inward  from  the 
base,  the  whole  pressure  on  the  base  is  more  than  the  weight 
of  the  liquid  in  the  vessel.  Explain  how  this  can  be. 

The  general  rule  for  the  total  pressure  on  a  horizontal 
surface  under  a  liquid,  like  the  base  of  the  vessel  in  either 
of  the  cases  just  given,  is  as  follows: 

RULE. — The  TOTAL  PRESSURE  on  a  horizontal  surface  below 
a  liquid  is  equal  to  the  area  times  the  depth  times  the  weight 
of  unit  volume  of  the  liquid. 

(6)  Whole  Pressure  on  a  Vertical  Wall. — If  the  surface  is 
vertically  placed  in  the  liquid  we  may  find  the  total  pres- 
sure on  it  as  follows:  Find 
the  average  pressure  on  the 
surface  and  multiply  this  by 
the  total  area.  Pressure  in- 
creases with  depth,  varying 
from  zero  at  the  top  of  a  sur- 
face placed  as  A,  in  Fig.  198, 
to  a  maximum  value  equal  to 

depth  X  density  at  the  bottom.  The  average  pressure  for  a 
rectangular  surface  is  therefore  the  pressure  at  a  point  half 
way  from  the  upper  edge  of  the  rectangle  to  the  lower 
edge.  The  student  may  work  out  for  himself  the  average 
pressure  and  area  for  the  other  cases  illustrated  in  Fig. 
198.  Where  the  surfaces  are  not  rectangles,  the  varying 
pressure  is  not  distributed  according  to  depth  alone,  and 
the  average  pressure  is  evidently  the  pressure  at  the  center 
of  gravity  of  the  surface.  It  is  evident  from  inspection  that 
this  statement  applies  equally  to  rectangles,  hence  for  sur- 
faces of  any  form  we  may  apply  the  following  general  rule : 


FIG.  198. 


ELEMENTARY  PRACTICAL  MECHANICS 


281 


The  total  pressure  on  ANY  PLANE  SURFACE  below  a  liquid 
is  equal  to  the  area  in  the  liquid  times  the  vertical  distance 
from  the  surface  of  the  liquid  to  the  center  of  gravity  of  the 
area  times  the  weight  of  unit  volume  of  the  liquid. 

(c)  Whole  Pressure  on  Surfaces  Inclined  at  an  Angle  with 
the  Vertical. — Liquid  pressure  is  always  perpendicular  to  the 
retaining  surface  (Article  154),  hence  the  general  rule  stated 
under  (6)  may  be  applied  to  these  surfaces  also. 

158.  Center  of  Pressure. — When  a  surface  is  immersed  in 
a  liquid,  every  point  on  the  surface  is  subjected  to  a  pres- 
sure at  right  angles  to  the  surface. 

All  these  pressures  constitute  a  system  of  parallel  forces 
(Fig.  199)  whose  resultant  may  be  found.  In  our  study  of 
liquid  pressure  we  have 
learned  how  to  find  the 
amount  of  this  resultant  in 
certain  cases,  but  we  have 
not  considered  the  point 
on  the  surface  at  which  it 
acts. 

This  point  is  called  the 

CENTER    OF    PRESSURE,    and 

is  defined  as  the  point  of 
action   of    the   single  force  FlG 

equivalent  to  the  whole  pres- 
sure exerted  by  a  fluid  on   any  plane  surface  with  which 
it  is  in  contact. 

If  the  plane  surface  is  horizontal,  the  center  of  pressure 
is  the  same  as  the  center  of  gravity  of  the  surface.  Thus, 
the  center  of  pressure  on  a  horizontal  rectangle,  circle,  tri- 
angle, etc.,  is  at  the  center  of  gravity  of  the  surface  and 
a  single  force  (equal  to  the  total  pressure  on  the  area) 
applied  at  that  point  will  produce  equilibrium. 

If,  however,  we  take  the  case  of  a  rectangular  surface 


282 


ELEMENTARY  PRACTICAL  MECHANICS 


placed  vertically  in  a  liquid,  since  the  pressures  increase 
proportionally  to  the  depth  and  the  total  force  on  the  sur- 
face depends  on  the  shape  and  position  of  the  surface,  the 
resultant  can  only  be  obtained  by  calculation  but  the  cen- 
ter of  pressure  is  always  below  the  center  of  gravity.  For  if 
the  pressures  on  equal  areas  on  different  parts  of  the  sur- 
face were  equal,  the  point  of  application  of  their  resultant, 
the  center  of  pressure,  would  coincide  with  the  center  of 
gravity.  But  the  pressure  increases  with  the  depth  and 
therefore  the  center  of  pressure  is  necessarily  below  the 
center  of  gravity.  The  exact  location  of  this  point  is 
found  to  be  as  follows: 

1.  With  a  rectangular  surface  whose  upper  edge  is  level 
with  the  liquid,  the  center  of  pressure  is  two-thirds  of  the 
way  from  the  top  on  the  line  joining  the  middle  points  of 
the  horizontal  sides. 

2.  With   a  triangular   surface   whose   base  is   horizontal, 
and  at  the  level  of  the  liquid,  the  center  of  pressure  is  at  the 
middle  of  the  line  which  joins  the  vertex  with  the  middle 
of  the  base. 

3.  With  a  triangular  surface  whose  vertex  is  level  with  the 
water,  the  center  of  pressure  is  in  the  line  joining  the  ver- 
tex and  the  middle  of  the  base,  and 
three-fourths  of   the  way  from  the 
vertex. 

159.  Stability  of  a  Retaining 
Wall.— Suppose  ABCD,  Fig.  200, 
represents  a  section  of  a  retaining 
wall.  The  resultant  water  pressure 
P  will  act  at  a  distance  equal  to 
J  the  depth  of  the  water  H  from 
the  bottom.  Its  tendency  to  overthrow  the  wall  by  tip- 
ping it  over  the  edge  at  A  will  be  measured  by  the 

TJ 

moment  PX— .     This  moment   will   be    opposed   by   the 


FIG.  200. 


ELEMENTARY  PRACTICAL  MECHANICS  283 

moment   of  the   weight   W  of  the  wall,   or  Wxd.     Evi- 
dently, in  order  that  the  wall  may  stand,  Wxd  must  be 

TT 

greater  than  PX— . 
o 


PROBLEMS 

1.  Find   the   whole   pressure   on   a   rectangular   surface 
6.0  ft.  by  4.0  ft.  immersed  vertically  in  water  with  the 
shorter  side  parallel  to  and  2.0  ft.  below  the  surface. 

2.  Find  the  whole  pressure  on  the  curved  surface  of  a 
vertical  cylinder  which  is  filled  with  a  liquid,  sp.  gr.  1.5, 
the  height  of  the  cylinder  being  20  cm.  and  the  radius  of 
the  base  7.0  cm. 

3.  A  flood  gate  is  6.0  ft.  wide  and  10  ft.  deep.     What  is 
the  total  pressure  on  the  gate  when  the  water  is  level  with 
the  top? 

4.  A  box  with  a  closed  top  is  30  cm.  long,  20  cm.  wide, 
and  18  cm.  deep.     Opening  into  the  top  of  the  box  is  a 
vertical  rectangular  tube  2X3  cm.  and  40  cm.  high.     Both 
box  and  tube  are  filled  with  water.     Find  the  following: 

(a)  The  pressure  and  the  total  pressure  on  the  bot- 
tom of  the  box. 

(b)  The  same  for  the  top  of  the  box. 

(c)  The  weight  of  the  water  in  box  and  tube  to- 
gether. 

(d)  Account  for  the  fact  that  the  total  pressure  on 
the  bottom  of  box  is  greater  than  the  whole  weight  of 
water  in  the  apparatus. 

(e)  Find  the  total  pressure  on  the  end  of  the  box. 

(f)  The  same  for  a  side. 

(g)  The  same  for  a  side  of  the  tube. 

5.  What  is  the  pressure  in  Ibs.  per  sq.in.  at  a  depth  of 
half  a  mile  in  sea-water?     Sp.  gr.  1.03. 

6.  A  cube  whose  edge  is  20  cm.  is  sunk  till  its  top,  which 
is  horizontal,  is  60  cm.  below  the  surface  of  water.     Find 
the  pressure  on  one  of  its  vertical  sides. 


284 


ELEMENTARY  PRACTICAL  MECHANICS 


7.  A  hole  6.0  ins.  square  is  made  in  a  ship's  bottom  20 
ft.  below  the  water  line.     What  force  is  required  to  hold 
a  piece  of  wood  tightly  over  the  hole? 

8.  A  mill  dam  is  40  ft.  long  and  the  water  is  15  ft.  deep, 
and  the  pond  is  500  ft.  long.     Find  the  force  urging  the 
dam  down  stream. 

9.  A  house  is  supplied  with  water  from  a  reservoir  240 
ft.  above  ground.     Find  the  pressure  per  square  inch  at  a 
tap  25  ft.  above  the  ground. 

10.  Canal  lock  is  60  ft.  long,  20  ft.  wide,  10  ft.  deep. 

(a)  Pressure  per  square  foot  on  bottom? 
(6)  Total  pressure  on  side? 

(c)  On  gate? 

(d)  Pressure  per  square  foot  on  gate  half-way  down? 

(e)  8.0  ft.  down? 

11.  ABC,  Fig.  201,  stands  in  water  with  BD  in  vertical 
line.     Apex  B  is  2.0  ft.  beneath  surface.     AD  =  DC  =  4.0  ft. 
BD=IQ  ft.     Compute  total  pressure  against  surface. 


FIG.  202 


12.  A  and  B  are  two  cylinders  placed  as  shown  in  Fig.  202. 
A  is  2.0  ft.  high  and  4.0  ins.  diameter.  B  is  2.0  ft.  high  and 
2.0  ft.  diameter.  When  both  are  filled  with  water  com- 
pute: 

(a)  Total  pressure  on  bottom  of  B. 

(b)  Pressure  on  scale  pan  supporting  them  if  cylin- 
ders weigh  25  Ibs. 


ELEMENTARY  PRACTICAL  MECHANICS  285 

13.  On  applying  a  pressure  gauge  to  a  water  pipe  of  a 
building  it  registers  75  Ibs.  per  sq.in.  How  high  is  the 
supply  reservoir  above  this  point? 

160.  Resultant  Upward  Force  on  Bodies  Immersed  in  a 
Fluid. — If  we  take  a  piece  of  iron  weighing  1  Ib.  arid  hang 
it  wholly  immersed  in  water  from  one  scale  of  a  balance, 
we  find  that  it  now  seems  to  weigh  less  than  1  Ib.  The  dif- 
ference in  the  weight  in  the  two  cases  is  due  to  the  resultant 
upward  pressure  of  the  water  on  the  iron. 

In  Fig.  203  we  have  a  rectangular  block  immersed  in 
water  with  one  surface  horizontal.      It  is  evident  that  the 
pressure  on  one  vertical  side  of  the  block  is  just  equal  to 
that  on  the  other  side,  and  that  there  will 
be  in  general  no  resultant  horizontal  pres- 
sure on  the  block.     The  resultant  upward 
pressure  is  easily  found  as  follows:     The 
pressure  on  the  top  of  the  block  is  evi- 


dently that  of  a  column  of  the  liquid  having         FIG.  203. 
AB  for  a  base  and  AE  for 'height,  while  the 
pressure  on  the  bottom  of  the  block  is  that  on  a  column  of 
the  liquid  having  DC  for  a  base  and  DE  for  height.     The 
resultant  upward  pressure  must  be  equal  to  the  difference 
between  the  weight  of  these  two  columns,  i.e.,  to  the  weight 
of  a  column  of  liquid  ABCD,  that  is,  to  the  weight  of  the 
liquid  displaced.     If  the  body  be  of  irregular  shape,  a  more 
general  proof  must  be  employed,   but  the  result  is  the 
same.     Hence, 

ARCHIMEDES'  PRINCIPLE.— A  body  immersed  in  a  liquid 
loses  a  weight  equal  to  the  weight  of  the  fluid  displaced. 

It  is  evident  that  Archimedes'  principle  furnishes  a  con- 
venient method  of  determining  the  Specific  Gravity  of  a 
body  and  also  the  volume  of  irregular  bodies.  Thus  suppose 
a  body  weighing  80  gms.  in  air,  weighs  only  60  gms.  when 


286 


ELEMENTARY  PRACTICAL  MECHANICS 


immersed  in  water.  The  loss  of  weight  is  20  gins,  and 
according  to  Archimedes'  Principle,  this  is  the  weight  of  a 
volume  of  water  equal  to  the  volume  of  the  body. 


Its  specific  gravity  = 


weight  of  body  _80 

weight  of  equal  vol.  water     20 


FIG.  204. 


And  since  1  cc.  of  water  weighs  1  gm.,  if  a  volume  of  water 
equal  to  that  of  the  body  weighs  20  gms.  the  volume  of  the 
body  is  20  cc. 

Archimedes'  Principle  applies  to  bodies  immersed  in  a 
gas  as  well  as. to  those  in  a  liquid.     Hence  the  real  weight 


ELEMENTARY  PRACTICAL  MECHANICS  287 

of  all  bodies  (that  is,  their  weight  in  a  vacuum)  is  greater 
than  their  weight  in  air,  which  may  be  called  their  appar- 
ent weight,  by  the  weight  of  the  air  displaced. 

Do  two  bodies  having  the  same  apparent  weight  have 
always  the  same  real  weight?  When  do  they?  Give  the 
rule  for  finding  the  real  weight  of  a  body  from  its  apparent 
weight  and  the  other  necessary  data.  Explain  the  lifting 
power  of  a  balloon. 

161.  Hydrometers. — Floating  bodies  sink  in  the  liquid 
until  they  displace  their  own  weight.  The  resultant  up- 
ward force  then  equals  the  downward  force  of  gravity  and 
the  bodies  are  in  equilibrium.  It  is  evident  that  a  floating 
body  must  sink  deeper  in  a  light  liquid  than  in  a  denser 
in  order  to  displace  its  own  weight;  or,  in  other  words,  the 
volume  of  two  liquids  displaced  must  be  inversely  as  the 
densities.  This  principle  is  the  basis  of  the  usual  forms  of 
HYDROMETERS,  or  instruments  which  indicate  the  density 
of  liquids  by  the  scale  readings  on  their  stems.  Common 
types  of  hydrometers  are  shown  in  Fig.  204.  These  instru- 
ments are  calibrated  by  placing  them  in  liquids  of  known 
density  and  marking  upon  the  stem  the  point  to  which 
they  sink,  that  for  pure  water  being  marked  1.000.  To 
provide  greater  sensitiveness,  a  single  instrument  is  de- 
signed to  measure  a  small  range  of  densities  only.  Special 
instruments  are  thus  made  for  testing  alcohol,  milk,  acids, 
etc. 


PROBLEMS 

1.  A  piece  of  glass  weighs  24  gms.  in  air  and  16  gins,  in 
water.     Find  its  specific  gravity. 

2.  A  block  of  wood  is  placed  in  a  vessel  just  full  of  water. 
It  floats  half  submerged  and   100  cc.   of  water  run  out. 
Find  the  weight,  volume,  and  specific  gravity  of  the  wood. 


288  ELEMENTARY  PRACTICAL  MECHANICS 

3.  A  barge  with  vertical  sides,  floating  in  fresh  water,  is 
30  ft.  long  and  20  ft.  wide.     An  elephant  is  driven  upon 
the  barge  and  the  barge  is  then  found  to  have  sunk  2  ins. 
Find  the  weight  of  the  elephant. 

4.  A  man  weighs  75  kgm.  and  his  volume,  exclusive  of 
his  head,  is  72,000  cc.     How  many  cubic  centimeters  of 
cork  (sp.  gr.  0.25)  are  required  to  keep  the  man  floating 
with  his  head  above  water? 

5.  A  body  weighs  300  gms.  in  air,  248  gms.  in  water,  and 
206  gms.  in  an  acid.     What  is  the  sp.  gr.  of  the  acid? 

6.  A  piece  of  wood  weighs  120  gms.  in  air.     A  piece  of 
lead  weighs  30  gms.  in  water.     Both  together  weigh  20  gms. 
in  water.     Find  the  sp.  gr.  of  the  wood. 

7.  What  fractional  part  of  an  iceberg  (sp.  gy.  0.918)  will 
float  above  sea-water  (sp.  gr.  1.03)? 

8.  What  will  be  the  loss  of  weight  in   water  of  2  cu.ft. 
of  stone?     Of  2  cu.ft.  of  iron?     Of  2  cu.ft.  of  wood? 

9.  Find  the  volume  of  a  body  that  weighs  350  gms.  in  a 
vacuum  and  225  gms.  in  water. 

10.  A  piece  of  metal  weighs  36.0  Ibs.  in  air  and  32.0  Ibs. 
in  fresh  water.     What  will  it  weigh  in  sea-water  of  den- 
sity 1.03? 

11.  A  cylindrical  post  30  ft.  long  and  8.0  ins.  in  diam- 
eter, has  a  weight  of  880  Ibs.     It  has  6.0  ft.  of  its  length 
projecting  vertically  out  of  water.     What  force  is  required 
and  in  what  direction,  to  keep  it  in  this  position? 

12.  A  glass  tube  100  cms.  long  holds  210  gms.  of  mer- 
cury.    Compute  internal  diameter  of  the  tube. 

13.  An  iron  ball  weighing  105.  gms.  is  suspended  under 
water.     Tension  in  suspending  string? 

14.  An  iron  weight  of  1.0  kilo  is  to  be  a  cylinder  5.0  cms. 
thick.     What  diameter  must  it  be  made? 


ELEMENTARY  PRACTICAL  MECHANICS  289 

GASES 

162.  Gases  Compared  with  Liquids. — Gases,  like  liquids, 
have  weight  and  such  perfect  freedom  from  molecular  fric- 
tion that  they  obey  the  laws  of  Pascal  and  Archimedes. 

Gases,   however,  differ  from  liquids  in  three  important 
respects: 

(1)  They  behave  as  if  their  molecules  repelled  each 
other. 

(2)  They  are  very  compressible. 

(3)  They  expand  much  more  rapidly  when  heated. 
The  proof  of  the  first  of  these  statements  is  seen  in  all 

the  familiar  cases  where  a  gas  expands,  as  its  pressure  is 
reduced.  A  gas  must  be  kept  in  a  vessel  enclosed  on  all 
sides  or  it  will  expand  indefinitely. 

The  proof  of  (2)  is  also  given  by  many  common  experi- 
ments, some  of  which  will  presently  be  considered. 

163.  The  Density  of  a  Gas  Depends  upon  Both  its  Pres- 
sure and  its  Temperature. — If  no  specifications  are  given, 
it  may  usually  be  assumed  that  Normal  Conditions  are  in- 
tended, i.e.,  a  pressure  of  30  ins.  or  76  cm.  of  mercury  and 
a    temperature    of    0°    Centigrade,    32°    Fahrenheit.     The 
specific  gravity  of  a  gas  is  usually  referred  to  that  of  dry 
air  as  a  standard. 

164.  Atmospheric  Pressure. — That  the  air  has  weight  is 
a  fact  not  known  till  about  1620,  when  the  work  of  Galileo, 
Torricelli,  Pascal,  Guericke,  and  others  established  the  fact. 
The  weight  of  the  air  may  be  demonstrated  in  the  same 
way  as  the  weight  of  anything  else,  that  is,  by  weighing  a 
vessel  full  of  air  and  then  the  same  vessel  empty,  and  ob- 
serving the  difference.     The  weight  of  the  air  has  the  most 
important  consequences  upon  the  conditions  not  only  of 
human  life,  but  also  upon  the  conduct  of  very  many  prac- 
tical operations  and  scientific  experiments.     All  the  phe- 


290 


ELEMENTARY  PRACTICAL  MECHANICS 


nomena  commonly  spoken  of  under  the 
term  "  suction/'  are  due  to  inequalities 
of  atmospheric  pressure,  i.e.,  the  rise 
of  water  from  the  cistern  into  the  bar- 
rel of  a  common  pump.  The  operation 
of  siphons,  of  various  types  of  air- 
pumps,  etc.,  depends  upon  atmospheric 
pressure. 

165.  The  Barometer.—  Torricelli  (an 
Italian,  died  1647)  devised  an  experi- 
ment to  show  the  weight  of  the  air, 
which  may  be  performed  as  follows: 
Take  a  glass  tube  at  least  34  ins.  in 
length,  open  at  one  end  and  closed  at 
the  other.  Fill  it  carefully  with  mer- 
cury and,  placing  the  thumb  over  the 
open  end,  invert  the  tube  with  this 
end  under  the  surface  of  some  mer- 
cury contained  in  a  cup.  On  removing 
the  thumb,  the  mercury  will  be  found 
to  sink  somewhat  and  will  soon  come 
to  rest  with  its  surface  at  a  level 
about  30  ins.  higher  than  the  level 
of  the  mercury  in  the  cup.  Now  it  is 
evident  that  the  pressure  of  the  col- 
umn of  mercury  in  the  tube  is  equal 
to  the  height  of  the  column  times  the 
weight  of  a  unit  volume  of  the  mer- 
cury, or  about  15  Ibs.  per  sq.in.  The 
atmospheric  pressure  on  the  surface  of 

FIG.  205.  —  Standard    the  mercury  in  the  cup  therefore  must 
barometer.  alg()  be  ab()ut  15  lbg>  per  gq>in>  f() 


tain  the  mercury  in  the  tube.     And  since  the  pressure  at 
the  mouth  of  the  tube  must  always  be  the  same  in  all 


ELEMENTARY  PRACTICAL  MECHANICS 


291 


directions,  if  atmospheric  pressure  increases,  the  mercury 
must  rise  in  the  tube  until  its  weight  becomes  sufficient 
for  equilibrium;  and  if  atmospheric  pressure  decreases,  the 
column  of  mercury  must  sink  for  a  similar  reason.  Torri- 
celli's  apparatus  is  therefore  a  simple  BAROMETER,  or 
instrument  for  measuring  atmospheric  pressure. 

The  common  form  of  standard  barometer  is  shown  in  Fig. 
205.  As  the  scale  for  reading  the  height  of  the  column  of  mer- 
cury is  rigidly  attached  to 
the  case,  the  first  operation 
in  determining  atmospheric 
pressure  with  this  form,  is 
to  set  the  surface  of  the 
mercury  in  the  cistern  at 
zero  of  the  scale.  This  is 
done  by  means  of  a  screw, 
which  raises  or  lowers  the 
flexible  bottom  of  the  cis- 
tern until  the  mercury  is 
just  in  contact  with  the  tip 
of  the  ivory  point  which 
marks  the  zero  of  the  scale. 

The  so-called  Aneroid 
Barometer,  Fig.  206,  consists 
of  a  thin  metallic  box  tightly  closed  on  all  sides,  and  having 
a  top  of  corrugated  form  which  is  easily  compressed  to  a 
slight  degree.  The  air  in  the  box  is  slightly  exhausted  so 
that  the  variations  of  the  pressure  of  the  air  outside  cause 
the  top  of  the  box  to  move  in  or  out  through  small  dis- 
tances. A  suitable  set  of  small  gears  and  levers  transmits 
these  motions  to  a  pointer  moving  over  a  scale,  The 
scale  is  marked  off  in  the  first  place  by  comparing  the 
aneroid  with  a  standard  mercury  barometer,  and  the  instru- 
ment requires  to  be  frequently  so  compared  to  be  reliable. 


FIG.  206.— Aneroid  barometer. 


292  ELEMENTARY  PRACTICAL  MECHANICS 

166.  Standard  Pressure. — As  has  been  stated,  the  mer- 
cury will  stand  in  a  barometer  tube  at  a  height  of  about 
30  ins.  or  76  cms.  This  is,  however,  subject  to  considerable 
variation  due  to  the  changes  in  weather,  etc.  "  Standard 
'Pressure  "  means  a  pressure  of  30  ins.,  or  what  is  almost 
exactly  the  same,  76  cms.  of  mercury  at  a  temperature  of 
0°  C.  This  is  the  pressure  to  be  assumed  in  the  absence  of 
any  specifications,  and  is  the  pressure  to  which  all  scien- 
tific data  are  reduced  in  tables,  etc. 

Pressures  expressed  in  heights  of  the  column  of  mercury 
which  atmospheric  pressure  will  support  may  readily  be 
reduced  to  pressures  in  Ibs.  per  sq.  in.,  or  grams  per  sq.  cm. 
by  multiplying  height  by  the  density  expressed  in  proper 
units.  (See  Article  152.)  Thus,  30  ins.  of  mercury  corre- 
sponds to  14.7  Ibs.  per  sq.  in.  (in  round  numbers,  15  Ibs.  per 
sq.  in.). 

PROBLEMS 

1.  What  fractional  part  of  the  air  has  been  removed  from 
a  vessel  when  the  pressure  has  fallen  from  74  to  25  cms.? 

2.  Barometer  stands  at  29  ins.     Compute  atmospheric 
pressure  in  pounds  per  square  inch. 

3.  Atmospheric  pressure  sustains  30.2  ins.  of  mercury. 
How  high  a  column  of  water  will  it  sustain? 

4.  A  pair  of  Magdeburg  hemispheres  are  6.0  ins.  in  diam- 
eter.    Pressure  inside  is  .50  in.  of  mercury,  outside  30  ins. 
Compute  force  required  to  pull  them  apart. 

5.  If  in  ascending  a  mountain  the  barometer  falls  from 
76  to  51  cms.,  find  the  decrease  in  total  pressure  on  an  area 
of  1  sq.ft. 

6.  If  a  man's  body  has  a  surface  of  18  sq.ft.,  find  the 
total  pressure  upon  him  in  tons. 

7.  Find  the  specific  gravity  of  air  when  the  barometer  is 
at  58  cms.,  if  it  is  .0013  when  the  barometer  is  at  76  cms. 

8.  -Find  the  greatest  height  that  water  can  be  carried 


ELEMENTARY  PRACTICAL  MECHANICS  293 

over  a  siphon  at  the  top  of  a  mountain  where  the  barom- 
eter stands  at  58  cms. 

9.  A  hollow  sphere  has  an  external  diameter  of  6.0  ins. 
What  is  the  total  force  due  to  atmospheric  pressure  hold- 
ing any  two  hemispheres  together?     Answer  in  pounds.. 

10.  The  piston  of  a  steam  engine  has  a  diameter  of  10 
ins.,  and  the  steam  exerts  a  pressure  upon.it  of  5  atmos- 
pheres.    Find  the  effective  force  when  the  other  side  of  the 
piston  is  exposed  to  the  atmosphere. 

11.  If  the  density  of  air,  like  that  of  water,  were  uniform 
and  equal  to  that  of  the  air  at  the  sea  level,  how  high 
would  the  atmosphere  extend?     Assume  the  height  of  a 
water  barometer  to  be  34  ft. 

167.  Boyle's  Law. — We  have  seen  that  the  characteristic 
qualities  of  a  gas  are  its  expansibility  and  compressibility. 

Experiment  shows  that  if  the  volume  of  a  gas  be  doubled, 
its  pressure  is  reduced  one-half;  if  the  volume  be  made  one- 
third  as  great,  the  pressiire  will  become  three  times  as  great, 
etc.  In  general,  we  have  the  statement  known  as 

BOYLE'S  LAW. — The  volume  of  a  gas  varies  inversely  as 
its  pressure,  when  the  temperature  remains  constant.  (This 
is  sometimes  called  Marriotte's  Law.) 

As  the  mass  of  the  gas  remains  the  same,  it  follows  that 
the  density  of  a  gas  must  increase  as  the  volume  decreases, 
and  vice  versa.  Hence  Boyle's  Law  may  also  be  stated  thus: 

The  pressure  of  a  gas  is  proportional  to  its  density,  the 
temperature  remaining  constant.  Or, 

The  weight  of  gas  in  a  given  volume  at  a  constant  tempera- 
ture is  directly  proportional  to  its  pressure. 

To  state  this  law  in  symbols,  let  V\  and  V2  be  the  vol- 
umes at  pressures  PI  and  P2  respectively;  also  let  di  and 
d2  be  the  corresponding  densities.  Then 

V\     F*2  d\     PI 

=         or    Pi7i=P2^2,     also        =--. 


294  ELEMENTARY  PRACTICAL  MECHANICS 

jv 

For  a  given   constant  volume   with  varying   pressures, 

Wi^Pi 

W2    P* 

It  should  be  added  that  the  laws  stated  above  are  not 
absolutely  obeyed  by  any  gas.  The  variations,  are  however, 
so  small  as  to  be  discernable  only  by  the  most  delicate 
experiments  and  in  all  practical  work  and  most  scientific 
investigations  it  is  considered  that  all  gases  follow  Boyle's 
Law  precisely.  A  gas  which  obeys  Boyle's  Law  and  an- 
other law  known  as  Charles'  Law  (to  be  stated  under  the 
subject  of  Heat),  is  known  as  a  PERFECT  GAS.  The  idea  is 
of  value  in  theoretical  discussions.  As  just  stated,  in  prac- 
tical work  all  gases  are  assumed  to  be  Perfect  Gases,  though 
none  of  them  is  so  in  fact. 

168.  Dalton's  Law. — This  is  an  extension  of  Boyle's  Law 
for  a  mixture  of  different  gases.  If  several  gases  which  do 
not  act  chemically  on  each  other  are  placed  in  a  vessel,  the 
pressure  on  the  sides  of  the  vessel  is  the  sum  of  the  pres- 
sures due  to  the  different  gases.  Thus  if  the  pressures  of 
the  different  gases  are  PI,  P2,  PS,  P±,  etc.,  the  intensity  of 
the  total  pressure  exerted  by  the  mixture  is  Pi  +  P2  +  Pa  + 
P4,  etc. 

Dalton's  Law  may  be  stated  as  follows:  When  a  mixture 
of  several  gases  at  the  same  temperature  is  contained  in  a 
vessel ,  each  gas  produces  the  same  pressure  as  if  the  others 
were  not  present. 

PROBLEMS 

1;  A  quantity  of  air  at-0°  C.  and  under  a  pressure  of  95 
cms.  measures  100  cc.  Find  its  volume  at  76  cms.  pressure. 

2.  What  will  1  liter  of  air  weigh  at  0°  C.  under  a  pres- 
sure of  4  atmospheres,  if  1  liter  of  air  at  0°  C.  and  76  cms. 
pressure  weigh  1.29  gms.? 


ELEMENTARY  PRACTICAL  MECHANICS  295 

3.  A  gas  inclosed  in  a  cylinder  with  a  freely  moving  pis- 
ton has  a  volume  of  1.23  cu.ft.  when  the  barometer  is  at 
29.5  ins.     To  what  height  must  the  air  pressure  fall  in 
order  that  the  volume  of  the  gas  may  become  1.34  cu.ft. 
at  the  same  temperature? 

4.  What  fractional  part  of  the  air  in  the  receiver  of  an 
air-pump  has  been  removed  when  the  pressure  within  has 
fallen  from  30  to  5.6  ins.? 

5.  A  diving  bell  is  lowered  till  its  lower  edge  is  at  depth 
of  15  ft.  below  the  surface  of  the  water.     The  barometer 
pressure   at   the   time    is   30  ins.      The   space   within  the 
cylindrical  bell  is   12  ft.  high.      At  what  point  within  the 
bell  will  the  water  level  stand? 

6.  If  a  room  contains  23  Ibs.  of  air  at  a  pressure  of  75 
cms.,  how  many  pounds  will  it  contain  at  a  pressure  of 
72  cms.? 

7.  A  cylinder  40  cms.  long  is  fitted  with  a  piston  2.0  cms. 
thick.     The  piston  is  placed  at  the  center  of  the  cylinder 
and  both  sides  of  it  are  opened  to  the  air  and  then  closed. 
The  barometer  stands  at  74  cms.     The  piston  is  then  forced 
to  a  position  such  that  its  center  is  9.0  cms.  from  one  end 
of  the  cylinder.     What  is  then  the  pressure  in  each  side? 

8.  The  mouth  of  vertical  cylinder  18  ins.  high  filled  with 
air  is  closed  by  piston  weighing  6.0  Ibs.  and  area  6.0  sq.ins. 
If  the  piston  is  allowed  to  fall,  how  far  will  it  descend? 

9.  1000  cc.   of  gas  at  760  mm.  pressure  will  have  what 
volume  under  500  mm.?     Under  1800  mm.? 

10.  100  cu.ft.  of  air  at  pressure  of  15  Ibs.  per  sq.in.  is 
compressed  to  36  cu.ft.  at  the  same  temperature.     Pressure 
then? 

11.  A  cylinder  40  cms.  long  inside  is  fitted  with  a  tight 
piston.     When  the  gas  inside  is  at  atmospheric  pressure, 
75  cms.,  the  inner  face  of  the  piston  is  35  cms.  from  the 
closed  end.     The  piston  is  pushed  in  so  that  the  space 
within  is  successively  30,  25,  20,  15,  10,  and  5  cms.  long. 
Calculate  the  pressures   at   these   positions   and  plot   the 


296 


ELEMENTARY  PRACTICAL  MECHANICS 


results  on  a  pressure-volume  diagram.     (Use  volumes  as 
abscissae.) 

12.  Plot  a  pressure  volume-diagram  from  the  following 
data,  using  pressures  as  ordinates  and  volumes  as  ab- 
scissae. 


Pressures. 
Lbs.  per  Sq.  In. 

Volume. 
Cu.  In. 

Pressures. 
Lbs.  per  Sq.  In. 

Volumes. 
Cu.  In. 

200 

100 

100 

200 

180 

111 

80 

250 

160 

125 

60 

234 

140 

143 

40 

500 

120 

167 

20 

1000 

169.  Measurement  of  Gas  Pressures. — Manometers  or 
pressure  gauges  are  instruments  for  measuring  the  pres- 
sure of  gases  or  vapors  in  closed  systems.  The  ordinary 
commercial  form  of  vacuum  and  pressure  gauges  is  shown 
in  Fig.  207. 


FIG.  207. — Common  type  of  pressure  gauge. 

Two  forms  of  simple  manometers,  the  open  and  the 
closed  tube  types,  are  shown  in  Fig.  208. 

In  use,  type  one  is  rilled  with  some  liquid  of  known 
density,  and  the  end  a  is  connected  with  the  closed  ves- 


ELEMENTARY  PRACTICAL  MECHANICS 


297 


sel  or  pipe,  the  pressure  within  which  is  to  be  determined. 
The  difference  in  level  of  the  liquid  in  the  two  arms  indi- 
cates the  difference  between  the  pressure  exerted  by  the 
atmosphere  and  the  pressure  exerted  by  the  confined  fluid 
within  the  vessel.  It  may  therefore  be  used  for  pres- 
sures either  greater  or  less  than  atmospheric  pressure. 
The  actual  pressure  is  obtained  by  computation  from  the 
barometer  reading  and  the  observed  difference  in  reading 
of  the  manometer  columns. 

It  should  be  carefully  noted  that  the  difference  between 
the  levels  in  the  manometer  gives,  when  reduced  to  pres- 
sure units,  the  EFFECTIVE  PRESSURE  within  the  vessel,  and 
to  this  must  be  added  the  atmospheric  pressure  to  obtain 
the  TOTAL  PRESSURE.  It  is  the  former  of  these  that  is 
given  by  the  ordinary  steam  gauge  as  used  on  boilers,  etc. 

Type  II  is  used  for  the  measurement  of  higher  pressures. 
The  end  b  is  closed  and  contains  a  column  of  air  above 
the  mercury  in  the  tube. 
In  using  such  a  manom- 
eter, readings  are  first 
taken  of  the  volume  of 
air  (usually  expressed 
simply  as  lengths  of  tube 
occupied  by  the  air),  and 
the  difference  of  levels  of 
the  mercury  when  the  end 
a  is  open  to  the  air.  Con-  FIG.  208. 

nection  is  then  made  at  a 

with  closed  vessel.  The  pressure  of  the  fluid  in  this  vessel 
will  change  the  level  of  the  mercury.  Readings  are  then 
taken  of  the  volume  of  the  air  and  the  difference  of  levels  of 
the  mercury  in  the  tubes.  From  the  two  sets  of  readings  the 
volume  of  the  air  and  the  pressure  upon  it  (the  reading  of 
the  barometer  being  known)  in  the  beginning,  and  the 


298  ELEMENTARY  PRACTICAL  MECHANICS 

volume  of  the  air  under  the  unknown  pressure  of  the  vessel 
are  known.  The  unknown  pressure  may  then  be  com- 
puted by  the  application  of  Boyle's  Law.  This,  corrected 
for  the  difference  of  level  of  the  mercury  r  will  give  the 
pressure  of  the  closed  vessel. 


PROBLEMS 

1.  The  pressure  of  gas  in  a  main  is  found  to  cause  the 
water  in  one  arm  of  an  open-arm  manometer  to  stand  2.4 
ins.  higher  than  in  the  other.     Find  the  pressure  of  the  gas 
in  pounds  per  square  inch. 

2.  An  open-arm  manometer  is  filled  with  mercury  and 
when  it  is  connected  to  a  certain  cylinder  of  gas  the  mer- 
cury in  the  open  arm  rises  to  a  position  4  cms.  above  the 
level  of  the  mercury  in  the  closed  arm.     Find  the  pressure 
indicated  in  kilograms  per  square  centimeter. 

3.  In  the  preceding  problem,  what  is  the  total  pressure 
if  the  barometer  at  the  time  stands  at  73  cms.? 

4.  In  the  diagram  of  the  closed-arm  manometer  on  the 
preceding  page,  assume  the  following  readings:    Before  the 
connection  was  made  to  the  vessel  the  level  d  was  4.0  cms. 
higher  than  the  level  c,  and  the  barometer  stood  at  75  cms. 
The  air  column  was  then  10  cms.  long.     After  the  connec- 
tion was  made  to  the  vessel,  the  mercury  stood  2.3  cms. 
higher  at  c  than  at  d,  and  the  air  column  was  6.1  cms.  long. 
Find  the  pressure  of  the  fluid  in  the  vessel  (a)  in  centi- 
meters  of   mercury;     (b)   in  atmospheres;     (c)   in   feet   of 
water;    (d)  in  pounds  per  square  inch. 

5.  Data  similar  to  the  preceding  problem  to  find  the  same 
items.     Before  connection,  d  is  5.2  cms.  higher  than  c;   air 
column  15  cms. 

6.  I  wish  to  measure  the  water  pressure  at  a  tap  and 
connect  with  an  open  mercury  manometer.     When  the  tap 
is  opened,  mercury  is  forced  to  height  of  110  cms.  in  the  open 
arm.     Express  "  head  "  in  feet. 


APPENDIX 


USEFUL  NUMBERS 
circumference 


diameter 
7T2 =9.8696;    --  =  .3183. 

7T 

nd2 
Area  of  circle  =;rr2= —  =.7854d2. 

•      4 

Surface  of  cylinder  =  2nrl  +  27rr2. 
Volume  of  cylinder—  nr2l. 
Surface  of  sphere =47rr2. 

Trd3       47T713 

Volume  of  sphere =— -  =  — — . 
6         3 


METRIC-ENGLISH  EQUIVALENTS 

1  cm.       =     .39  in.  1  in.       =     2.54  cms. 

1  m.         =39.37  ins.  1  ft.        =  30.48  cms. 

1m.         =  3.23  ft.  1  ft.        =       .305  m. 

1  km.       =     .6  mile  1  mile    =     1.60  km. 

1  gm.       =       .035  oz.  (avoir.)  1  oz.       =  28.35  gms, 

1  kgm.     =     2.204  Ibs.  (avoir.)  1  Ib.       =453.6  gms. 

1  sq.  cm.  =     .154  sq.  in.  Isq.  in.  =     6.45  sq.  cms. 

1  cu.  cm.  =     .061  cu.  in.  1  cu.  in.  =   16.39  cu.  cms. 

299 


300  APPENDIX 

WEIGHTS  (APPROXIMATE) 

1  cu.  ft.  of  water  weighs 62.5  Ibs. 

1  cu.  ft.  of  water  weighs 1000.  ozs. 

1  cu.  in.  of  water  weighs .036  Ib. 

1  cu.  ft.  of  air  ("  Standard  ")  weighs 0817  Ib. 

1  cu.  cm.  of  water  weighs 1.0  gm. 

1  cu.  cm.  of  air  ("  Standard  ")  weighs  ....  .00129  gm. 

1  gallon  (231  cu.  ins.)  water  weighs 8.32  Ibs. 

1  cu.  in.  cast  iron  weighs .26  Ib. 

1  cu.  ft.  cast  iron  weighs 450.      Ibs. 

1  cu.  in.  brass  weighs .30  Ib. 

1  cu.  in.  copper  weighs. .32  Ib. 

1  cu.  in.  lead  weighs .41  Ib. 

1  cu.  ft.  sandstone  weighs 144.      Ibs. 

1  cu.  ft.  slate  weighs 175.      Ibs. 

1  cu.  in.  mercury  weighs .49  Ib. 

UNITS  OF  FORCE,  WORK,  POWER,  ETC. 

Value  of  "  g  "  at  New  York 980.2    cm/sec2 

Value  of  "  g  "  at  New  York 32.16  ft/sec2 

(N.B. — 980  cms.  or  32  ft.  are  close  enough  for  ordinary  use.) 

1  dyne  =   .00102  gm. 

1  poundal        =1.3825  dynes. 
Ift.-lb.  =  1.356  X107  ergs. 

1  joule  =107  ergs. 

•  1  horse-power  =  33,000  ft-lbs/min. 
1  horse-power  =      550  ft-lbs/sec. 
1  horse-power  =  7.46  X 109  ergs/sec 
1  horse-power  =  746  watts. 
1  watt  =   .00134  horse-power. 

1  watt  =  107  ergs/sec.  =  1  joule/sec. 


APPENDIX  301 

MECHANICAL  EQUIVALENTS  OF  HEAT 

1  gm.  of  water  heated  1°  C.=4.2X107  ergs. 
1  Ib.    of  water  heated  1°  C.  =  1400  ft.-lbs. 
1  Ib.    of  water  heated  1°  F.  -778  ft.-lbs. 

The  combustion  of  1  Ib.  of  coal  produces  about  14,000 
B.T.U. 


COEFFICIENT  OF  ELASTICITY  (TENSION  AND  COMPRES- 
SION) 

Steel 30,000,000  lbs/in2. 

Wrought  iron 30,000,000  lbs/in2. 

Copper 15,000,000  lbs/in2. 

(Note. — For  torsion  use  I  of  above.) 

TENSILE  STRENGTHS 

(N.B. — These  values  are  to  be  regarded  as  only  averages; 
there  are  wide  variations  according  to  hardness,  composi- 
tion, etc.) 

Steel 100,000  lbs/in2. 

Wrought  iron 55,000  lbs/in2. 

Copper  (wire) 55,000  lbs/in2. 

Brass  (wire) 50,000  lbs/in2. 

FACTORS  OF  SAFETY  (MERRIMAN) 

For  Steady          For  Varying          *?„„  «•      i  » 

Material.  Load  Load  FM    v/     ^? 

(Buildings).  (Bridges).  (Machines). 

Timber 8  10  15 

Brick  and  Stone 15  25  35 

Cast  Iron 6  15  20 

Wrought  Iron 4  6  10 

Steel 5  7  15 


302  APPENDIX 

SIGNIFICANT  FIGURES 

The  results  of  all  experimental  work  should  be  so  ex- 
pressed as  to  indicate  as  nearly  as  possible  the  degree  of 
precision  with  which  the  work  was  performed.  It  is  evi- 
dent that  all  numbers  that  are  obtained  as  the  result  of 
measurements  are  limited  in  precision  by  the  nature  of 
the  apparatus  employed,  by  the  care  used  by  the  observer, 
the  size  of  the  units,  etc.  In  this  respect,  then,  such  quan- 
tities are  quite  different  from  the  pure  numbers  of  absolute 
value  as  employed  in  ordinary  arithmetical  operations. 

The  student  must  observe  carefully  the  following  rules 
in  all  laboratory  work  and  in  reports. 

I.  RECORDING  READINGS 

In  general,  scales,  etc.,  are  to  be  read  to  tenths  of  the 
smallest  divisions  marked  on  the  instrument.  The  last 
figure  entered  in  the  record  is  thus  assumed  always  to  be 
an  estimation  and  therefore  doubtful. 

Example  1. — 15.57  cms.  means  that  a  distance  was  meas- 
ured by  a  scale  subdivided  to  millimeters,  and  that  the 
observer  estimated  the  seven;  thus  the  distance  is  known 
to  be  between  15.5  and  15.6,  and  estimated  to  be  -j^ths 
of  the  way  between  these  two  values.  It  is  misleading, 
and  furnishes  only  a  clue  to  what  we  actually  know  about 
this  distance  to  record  it  as  15.6  or  15.570  cm. 

Example  #.— A  distance  is  being  measured  withrar  Tute 
subdivided  to  tenths  of  inches.  The  observer  finds  the  dis- 
tance to  be  as  nearly  exactly  seven  inches  as  he  can  dis- 
tinguish. This  should  be  recorded  7.00  in.  (not  7.0  or  7 
ins.).  Why? 

Example  3. — A  balance  is  capable  of  weighing  an  object 
to  .01  gm.  and  .001  gm.  can  be  estimated.  Notice  the 
correct  records  for  following: 


APPENDIX  303 

Eight  gms.  ...  .........  ...    8.000  gms. 

Eight  and  \  gms  .......  ....    8.500  gms. 

Eight  and  ^  Q-  gms  .......  .  .    8.070  gms. 

Eight  and  Y^^  gms  ........    8.008  gms. 

Eight-tenths  gm  ..........  .      .800  gms. 

r 

In  general  a  series  of  readings  made  with  the  same  in- 
strument should  all  show  the  same  number  of  plaees  filled 
in  to  the  right  of  the  decimal  point  even  if  one  or  all  these 
places  are  zeros.  Why? 

It  is  often  convenient  to  express  in  decimal  form  read- 
ings taken  from  scales  divided  into  halves  of  units,  quar- 
ters, eighths,  etc.  In  all  such  cases,  retain  only  as  many 
places  in  the  decimal  as  correspond  approximately  to  the 
same  degree  of  precision  as  would  be  expressed  by  the 
fraction,  i.e.,  to  the  nearest  half  unit,  to  the  nearest  quar- 
ter, etc.  If  the  first  decimal  figure  rejected  is  5  or  greater, 
call  the  preceding  figure  one  larger  than  before. 

A  study  of  the  following  table  should  make  this  clear: 

i  =  .5  J   =.1  |  =  .4  §   =.7 

i  =  .3  Jg=.06  f  =  .6  |  =.8 

i  =  .3  ^=.03  f  =  .9  &  =.19 

I  =  .2  ^T  =  -02  Etc,  Etc. 


II.  USE  OF  DATA  IN  CALCULATIONS 

Wherever  the  figure  following  the  doubtful  (last  re- 
tained) figure  is  5  or  greater  than  5,  increase  the  doubtful 
figure  by  unity.  Thus,  if  but  three  figures  are  to  be  kept, 
15.75,  15.76,  15.77,  15.78,  and  15.79  would  all  be  entered 
15.8. 

Notice  especially  that  the  location  of  the  decimal  point 
has  nothing  to  do  with  significant  figures.  Thus,  275,  27.5, 


304  APPENDIX 

2.75,  .275,  .0275,  .00275,  etc.,  are  all  results  expressed  to 
some  degree  of  precision,  and  in  each  there  are  three  and 
only  three  significant  figures,  the  5  being  the  doubtful  figure 
in  each. 

Averages. — In  averaging  a  series  of  determinations,  in 
general,  retain  in  the  result  the  same  number  of  significant 
figures  as  in  any  one  item. 

But  if  a  large  number  of  items  closely  agreeing  with  each 
other  are  averaged,  the  result  may  contain  one  more  sig- 
nificant figure  than  any  item. 

Multiplication. — After  the  operation,  keep  in  the  result 
as  many  figures,  counting  from  the  left,  as  there  are  sig- 
nificant figures  in  the  factor  having  the  lesser  number  of 
significant  figures. 

Division. — In '  dividing  one  number  by  another,  keep 
in  the  quotient  as  many  figures  as  there  are  significant 
figures  in  the  number  having  the  lesser  number  of  signifi- 
cant figures.  Continue  the  divisions  only  far  enough  to 
determine  the  required  figures. 

Note  on  Multiplication  and  Division. — Ciphers  immediately  following 
the  decimal  point,  when  there  are  no  figures  to  the  left  of  the  point,  do 
not  count  as  significant.  Study  the  following  examples: 

(a)  15.75     X3.08       =48.5. 
(&)       .096   X   .096     =     .0092. 

(c)  .1523X   .00113-     .000172. 

(d)  720        X3.1         =2200. 

(e)  900        X800        =720000. 

In  (e)  only  the  first  cipher  is  significant.  It  is  necessary 
to  add  the  other  three  to  express  the  number  properly. 

(/)   325.6-5-72.5=4.49. 

(g)         .0007859  -r- 157  =  .00000506. 


APPENDIX 


305 


Use  of  Pure  Numbers,  Constants,  etc. — In  using  pure 
numbers  and  constants  such  as  (3.1416),  .7854,  etc.,  do 
not  employ  more  figures  than  there  are  significant  figures 
in  the  values  depending  on  experiment  which  are  used 
with  them  in  the  same  calculation.  Thus  if  the  diameter 
of  a  circle  is  measured  as  4.51,  the  area  is  4. 51X4. 51  X 
.785  =  15.9.  The  use  of  more  numbers  in  the  constant 
lengthens  the  computation  and  gives  no  better  result. 
Why? 


CURVES 

Coordinate  Axes. — The  position  of  a  point  in  space  may 
be  fixed  by  reference  to  two  known  straight  lines  inter- 
secting  at  right  angles  in 
the  same  plane  as  the  point 
(OX  and  OF  of  the   Fig. 
209) .     Such  lines  are  known 
as  coordinate  axes. 

The  horizontal  line  (OX) 
is  known  as  the  "  axis  of 
abscissce  "or  "  X  axis," 
the  vertical  line  (OY)  as 
the  "  axis  of  ordinates  "  or 
"  F  axis."  The  point  of 
intersection  0  is  called  the 
origin.  The  abscissa  of  a 
point  is  its  horizontal  distance  from  OF;  its  ordinate  is 
its  vertical  distance  from  OX.  These  given,  the  position 
of  the  point  is  determined.  Thus  P  is  that  point  which 
has  an  abscissa  of  3,  an  ordinate  of  5,  PI  the  point  which 
has  abscissa  of  11,  ordinate  of  8,  etc.  • 

For  convenience,   squared  or  "  cross-section "  paper  is 
used  for  work  of  this  kind 


P1 

P 

1 

X 

2 

1 

1 

5 

1 

0 

1 

•z 

FIG.  209. 


306 


APPENDIX 


Curves— A.  succession  of  related  points  may  be  connected 
by  a  smooth  line,  thus  constituting  a  "  curve"  Such 
curves  are  frequently  the  most  convenient  and  the  clear- 
est way  of  representing  a  physical  law,  corrections  for 
errors  of  apparatus,  etc.  Suppose,  for  example,  that  it  is 
desired  to  show  the  relation  between  the  stretch  of  a  wire 
and  the  stretching  loads  producing  it,  data  being  as  follows; 

Load.  Increase  in  Length. 

51bs 010  inch 

10   "    019     " 

15   " 030     " 

20   "    040     " 

25   " .051     " 

Taking  the  stretching  loads,  expressed  in  some  con- 
venient scale  of  lengths,  as  ordinates,  and  the  correspond- 
ing elongations  similarly 
expressed,  as  abscissae,  a 
series  of  points  may  be 
located  as  just  explained, 
and  through  these  a  smooth 
line  may  be  drawn.  In- 
spection of  the  curve  thus 
produced  (Fig.  210)  will 
show  at  a  glance  what 
could  be  obtained  from  the 


/ 

/ 

/ 

/ 

/ 

/ 

2 

.020              .OiO              .000.              .08 
Elongation  in  Inches 

figures   only   on   more   ex- 
pIG  210  tended  analysis.     The  law, 

"  Elongation  is  propor- 
tional to  the  load  applied,"  is  seen  immediately,  from  the 
nature  of  the  curve. 

Had  the  curve  turned  continuously  more  and  more 
toward  either  the  X  or  the  Y  axis,  showing  in  one  case  a 
progressive  increase,  in  the  other  a  progressive  decrease  in 


APPENDIX  307 

elongation  with  increase  of  load,  or  had,  at  any  time,  a 
sudden  change  from  the  conditions  which  had  previously 
existed  occurred,  these  factors  would  have  been  brought  to 
the  attention  as  quickly. 

When  also,  as  here,  the  great  majority  of  points  lie  along 
a  straight  line  (or,  as  in  some  cases,  along  a  smooth  curve), 
any  experimental  errors  of  measurement  (as  in  the  elonga- 
tions for  loads  of  10  and  25  Ibs.),  will  be  shown  at  once  by 
the  fact  that  these  points  lie  slightly  off  the  line.  In  all 
such  cases,  the  curve  should  be  drawn  as  nearly  as  may  be 
through  all  points,  and  leaving  as  many  points  on  one  side 
as  on  the  other. 

The  student  must  in  all  cases  use  his  judgment  in  draw- 
ing the  curve  and  consider  the  conditions  of  the  experi- 
ment and  the  general  physical  law  illustrated. 

It  is  not  necessary,  and  indeed  often  not  advisable,  that 
ordinates  and  abscissae  be  expressed  in  the  same  scale.  Of 
course,  for  the  same  curve  all  abscissae  must  be  in  one 
scale,  and  all  ordinates  in  one  scale.  In  general,  the  scale 
adopted  should  be  that  most  convenient  for  the  particular 
values  which  will  at  the  same  time  give  a  curve  as  large 
as  the  paper  will  permit. 

One,  two,  five,  or  ten  units  to  a  square  will  be  found  the 
best.  Avoid  the  use  of  three  or  seven  units  per  square, 
or  other  inconvenient  subdivisions. 


GENERAL  DIRECTIONS  FOR  CURVE  SHEETS 

(1)  The  curve  must  be  done  neatly  in  India  ink. 

(2)  Heavy  lines  one  inch  in  from  the  margin  on  the 
ruled  portion  are  to  be  taken  as  axes,  except  where  all  the 
paper  is  necessary  for  the  curve.     The  origin,  i.e.,  the  in- 
tersection of  vertical  and  horizontal  axes,  should  be  at  the 
lower   left-hand   corner.     The   paper   may   be   used    with 


308  APPENDIX 

either  longer  or  shorter  side  as  vertical  axis,  according  to 
needs  of  the  curve. 

(3)  The  scale  on    which    the    curve    is    plotted    should 
be  so   selected   as   to  make   the  -curve    as   large   as  pos- 
sible. 

(4)  Each  axis  should  be  marked  with  the  quantity  which 
it  represents,  and  with  the  unit  in  which  these  quantities 
are  expressed,  e.g.,   "  loads  in  pounds  per  square  inch," 
"  elongations  in  inches/'  etc.     These  titles  should  be  let- 
tered upon  the  ruled  paper  between  margin  and  axes. 

(5)  Each  half-inch  line  along  both  vertical  and  horizontal 
axes  should  be  marked  with  the  value  which  it  represents. 
No  other  figures  are  to  be  used  in  locating  the  curve. 

(6)  The  points  fixing  the  curve  are  to  be  located  by  a 
small  dot  around  which  is  drawrn  a  small  circle  with  a  pair 
of  dividers. 

(7)  The  curve  should  usually  be  a  smooth  line  drawn  as 
nearly  as  possible  through  all  points.     It  will  represent  the 
most  probable  value  of  the  observations,  and  any  single 
point  lying  at  a  distance  on  either  side  of  the  line  will 
usually  be  a  result  of  error  in  observations.     Of  course 
judgment  must  be  used  in  drawing  this  conclusion,  and 
the  conditions  of  the  experiment  and  the  nature  of  the 
related  quantities  of  the  curve  must  always  be  taken  into 
account. 

(8)  The  name  of  the  student  and  the  date  should  be 
placed  at  the  bottom  of  the  sheet,  at  the  right,  in  small 
letters. 

(9)  The  title  of  the  curve  should  be  stated  in  the  lower 
right-hand  portion  of  the  curve  sheet  unless  this  interferes 
with  the  curve;    in  which  case  the  lower  left-hand  or  the 
upper  right-hand  portion  should  be  used. 

(10)  If  more  than  one  curve  is  drawn  on  the  same  paper 
for  comparison,  etc.,  use  the  same  origin  and  scales  for  aLL 


APPENDIX 


309 


Distinguish  the  curves  by  the  title  printed  along  the  curve, 
or  by  lines  of  different  colors. 

(11)  All  titles,  explanations,  etc.,  must  be  in  lettering, 
and  no  handwriting  should  appear  upon  the  curve  sheet. 

THE  EQUATION  OF  A  STRAIGHT  LINE 

It  is  often  desired  to  find  the  equation  that  corresponds 
to  a  given  line  (straight  or  curved)  plotted  on  squared 
paper.  In  this  course  it  will  not  be  necessary  to  obtain 
the  equation  of  a  curved  line.  A  simple  method  for  the 
equation  of  a  straight  line  follows: 

Let  AB,  Fig.  211,  be  a  line  plotted  as  usual  on  the  axes 
OX  and  OF,  and  meeting  the  axis  of  Y  at  the  point  A. 
(If  the  line  as  first  drawn 
does  not  cut  the  axis  of 
Y  it  must  be  extended  till 
it  does  so.) 

At  the  point  A  draw  a 
line  parallel  to  the  axis  of 
X.  Choose  any  point  on 


TIG.  211. 


the  line  as  P%,  and  draw 

. 
its  ordmate  7/2-     #2  is  the 

abscissa  of  this   point.     We  desire  to  obtain  an  equation 
that  will  give  us  the  relation  between  the  abscissa  and 
the  ordinate  for  this  and  every  other  point  on  this  line. 
We  notice  first  that  the  ordinate  y%  equals  the  intercept 
OA  on  the  Y  axis,  plus  PzDz,  or 


Also, 


i/i=OA+PiDi, 


and  so  on  for  every  point  on  the  line. 


310  APPENDIX 

The  value  of  the  intercept  OA  may  now  be  read  from 
the  curve.  Suppose  in  the  given  case  OA=S.  Next  read 
from  the  curve  values  of  the  altitude  and  base  of  any  tri- 
angle whose  hypothenuse  is  some  part  of  the  line  AB.  These 
values  are  to  be  expressed  in  units  of  the  respective  scales 
used  in  plotting  X  and  Y  and  not  as  actual  lengths  in 
inches.  The  triangle  AP2D2  will  serve.  Suppose  P2D2=4: 

T>     T\ 

and  AD2  =  W  in  the  given   case.     Then   --  -  =  .4.     But 

AL/2 


=  X2,  therefore  P2D2  =  Ax2. 
If  we  had  used  other  triangles  we  should  have  obtained 
the  same  ratio  between  altitude  and  base,  and  thus, 


Or,  in  words,  we  may  now  say  that  any  ordinate  equals 
the  intercept  on  the  Y  axis  plus  .4  of  the  abscissa  for  the 
same  point.  Let  x  and  y  be  the  coordinates  of  any  point 
on  the  line  AB]  then 


which  is  the  equation  desired. 

P2D2 

The  ratio  —-—  is  sometimes  called  the  slope  of  the  line. 
AD2 

We  may  now  state  the  general  rule  as  follows: 
RULE.  —  The  equation  of  a  straight  line  is  formed  by 
putting  y  equal  to  the  intercept  on  the   axis  of  Y  plus 
the  slope  times  x.    If  intercept  =  a,  and  slope  (ratio)  =m, 
we  have, 

y=a  +  mx. 


APPENDIX  311 

NOTE. — The  student  will  notice  that  the  equation  just 
given  is  perfectly  general.  If  the  line  cuts  the  axis  of  Y 
below  the  origin,  the  intercept  will  be  a  negative  term  and 
the  equation  will  be  of  the  form  y=  —a  +  mx.  If  the  line 
slopes  so  that  an  increase  in  the  value  of  the  abscissa  causes 
a  decrease  in  the  value  of  the  ordinate,  then  m  will  be  a 
negative  quantity,  y  =  a  —  mx.  It  is  possible,  of  course, 
that  both  a  and  m  may  be  negative  at  the  same  time,  as 
y=—a  —  mx.  The  student  should  draw  and  consider  care- 
fully lines  to  illustrate  each  case. 

EXERCISES 
1.  Locate  following  points: 


Point, 
(a)   . 

Abscissa. 
,  5   

Ordinate. 
.      .    .     3 

(W   - 

7  

10 

(c) 

5  

8 

(d)   . 

0   

12 

(e) 

,  5  

5 

m 

.  9   . 

0 

2.  Measure  carefully  the  lengths  of  9  ins.,  7  ins.,  5  ins., 
3  ins.,  respectively  in  metric  units.     Make  each  measure- 
ment three  times,  using  different  parts  of  the  scale.      Why? 
Take  the  average  of  the  three  readings  and  plot  a  curve, 
using  inches  as  ordinates  and  the  corresponding  number 
of  centimeters  as  abscissa. 

The  curve  will  pass  through  the  origin  or  zero-point  of 
each  scale.     Why? 

From  your  curve  find  the  value  of  1  inch  in  centimeters. 
What  is  the  -true  value?     What  is  your  per  cent  of  error? 

3.  A  determination  of  the  relation  of  bending  of  beam  to 
load  gave  following  results: 


312  APPENDIX 

Loads.  Deflections. 

10  Ibs 0.05  inch. 

20   "    0.10     " 

30   "    0.15     " 

40   "    0.21     " 

50   "    0.25     " 

60   " 0.29     " 

70   "    0.35     " 

Plot  curve  showing  relation  of  deflection  to  load,  using 
loads  as  ordinates. 

4.  Plot  a  straight  line  such  that  it  shall  gain  3  units  of 
abscissae  for  every  unit  gained  as  ordinates. 

5.  A  determination  of  Boyle's   Law  gave  the  following 
data: 

Pressure.  Volume  of  Gas. 

82.1 12.03 

88.2 : 11.20 

96.2 10.26 

105.5 9.35 

118.9 8.31 

135.5 7.29 

160.1 6.17 

Plot  the  curve  of  above  values,  using  pressures  as  ordi- 
nates. 

6.  A  determination  showing  the  effect  of  length  of  a  beam 
upon  its  stiffness  gave  the  following  data: 

Length.  Deflection. 

10  inches .005  inch 

20      "       035     " 

30      "       120     " 

40      "       ...  .285     " 


APPENDIX  313 

Plot    curve    showing   relation    of    deflections   to   length, 
using  deflections  as  abscissa?. 

7.  The  area  of  a  circle  varies  with  square  of  its  diameter. 
Plot  a  curve  to  show  relation  of  area  to  diameter  in  circle 
whose  successive  diameters  are  1,  2,  3,  4,  5,    and  6.     Use 
diameters  as  ordinates. 

8.  What  is  the  equation  of  the  line  which  cuts  the  axis 
of  Y  at  a  point  3.4  above  the  origin  and  which  rises  5  units 
for  every  8  of  horizontal  distance? 

9.  If  the  intercept  of  a  line  is  .56  and  its  slope  is  2.58, 
what  is  its  equation? 

10.  A  line  has  two  points  whose  coordinates  are  (8.5) 
and  (3.1).     Plot  the  line  and  obtain  its  equation,    i 

11.  What  is  the  slope  of  a  line  if  its  intercept  is  4  and 
x  =  l2  when  i/  =  6.8? 

12.  The  slope  of  a  certain  line  is   .532,   and  it  passes 
through  the  origin.     What  is  the  ordinate  of  a  point  on 
this  line  whose  abscissa  is  2.3? 

13.  Write  the  equation  of  the  following  lines,     a  =  inter- 
cept and  m  =  slope. 

(a)  a  =  5,  m=     .13. 

(b)  a=  —5,  m=   3. 

(c)  a- -2.3,  m=     .70. 

(d)  a  =  3,  ra=-.68. 

(e)  a  =—2.5,  m=  —  .45. 

14.  Draw  a  sketch  to  show  the  general  character  of  each 
of  the  lines  described  in  Problem  13. 


314  APPENDIX 

SIMPLE  TRIGONOMETRIC  FUNCTIONS 

Let  ABC,  Fig.  212,  be  any  given  angle,  and  let  P  be  any 
point  on  the  line  BC.  From  P  draw  PN  perpendicular  to 
the  side  BA,  thus  forming  the  right-angled  triangle  PNB. 
Suppose  we  measure  BP  and 
find  it  to  be  10  inches  long  and 
PN  and  find  it  to  be  5.05  inches 
long.  Then 


"! 
BP        10  FIG.  212. 

Now  let  us  choose  another  point  on  BC,  as  P',  draw  the 
perpendicular  P'N'  and  measure  lines  BP'  and  PfNf.  If 
BP'  is  I  as  long  as  BP,  or  8  inches,  then  P'N'  will  be  £ 

as  long  as  PN,  or  5.05X4  =  4.04  inches.     The  ratio -  = 

BP 

4.04 

_! —  =  .505  or  the  same  as  that  of  PN  to  BP.     So  we  might 
8 

choose  any  point  on  BC  and  always  obtain  the  same  ratio 
of  the  perpendicular  to  the  hypothenuse  so  long  as  we  use 
the  same  angle  ABC. 

BN 

Similarly,   — -  gives   a   second    constant   ratio   for    the 


PN 

given  angle,  and  -—  a  third,  all  of  which  are  independent 
nN 

of  the  position  of  P  and  dependent  only  on  the  angle  ABC. 

PN 

If  the  angle  ABC  be  changed,  then  the  ratios  -777:?   e^c-> 

Jor 

will  have  new  values,  but  these  again  will  be  the  same, 
no  matter  where  P  is  taken  on  the  line  BC.  Each  angle 
thus  has  a  certain  number  of  constant  ratios  among  which 


APPENDIX  315 

are  the  three  here  given,  and  these  ratios  are  given  distin- 
guishing names. 

PN 

The  ratio  — —  is  called  the  sine  of  the  angle  ABC. 

BN 

The  ratio  — —  is  called  the  cosine  of  the  angle  ABC 

PN 

The  ratio  — -  is  called  the  tangent  of  the  angle  ABC. 
BN 

Definitions. — Let  B  be  an  angle  (not  the  right  fingle)  of 
a  right-angled  triangle.  The  sine  of  the  angle  B  is  the 
ratio  of  the  side  opposite  the  angle  to  the  hypothenuse  of 
the  triangle. 

The  cosine  of  the  angle  B  is  the  ratio  of  the  side  adja- 
cent the  angle  to  the  hypothenuse  of  the  triangle. 

The  tangent  of  the  angle  B  is  the  ratio  of  the  side  oppo- 
site to  the  side  adjacent. 

In  calculations,  sine,  cosine,  and  tangent  are  always 
written  for  brevity,  sin,  cos,  tan.  Thus,  sin  30°  =  .500; 
cos  45°  =  .707;  tan  50°  =  1.19. 

The  values  of  these  ratios  have  been  calculated  for  all 
angles,  and  are  given  in  what  are  called  tables  of  trigono- 
metric functions.  Such  tables,  with  the  values  carried  out 
to  three  decimals,  will  be  found  on  the  following  page.  See 
also  page  68. 


316 


APPENDIX 


TRIGONOMETRIC  FUNCTIONS 


A 

Sin. 

Cos. 

Tan. 

A 

Sin. 

Cos. 

Tan. 

0 

.000 

1.000 

.000 

1 

.017 

.999 

.017 

46 

.719 

.695 

1.04 

2 

.035 

.999 

.035 

47 

.731 

.682 

1  .07 

3 

.052 

.999 

.052 

48 

.743 

.669 

1.11 

4 

.070 

.998 

.070 

49 

.755 

.656 

1.15 

5 

.087 

.996 

.087 

50 

.766 

.643 

1.19 

6 

.105 

.995 

.105 

51 

.777 

.629 

1.23 

7 

.122 

.993 

.123 

52 

.788 

.616 

1.28 

8 

.139 

.990 

.141 

53 

.799 

.602 

1.33 

9 

.156 

.988 

.158 

54 

.809 

.588 

1.38 

10 

.174 

.985 

.176 

55 

.819 

.574 

1.43 

11 

.191 

.982 

.194 

56 

.829 

.559 

1.48 

12 

.208 

.978 

.213 

57 

.839 

.545 

1.54 

13 

.225 

.974 

.231 

58 

.848 

.530 

1.60 

14 

.242 

.970 

.249 

59 

.857 

.515 

1.66 

15 

.259 

.966 

.268 

60 

.866 

.500 

1.73 

16 

.276 

.961 

.287 

61 

.875 

.485 

1.80 

17 

.292 

.956 

.306 

62 

.883 

.469 

1.88 

18 

.309 

.951 

.325 

63 

.891 

.454 

1.96 

19 

.326 

.946 

.344 

64 

.898 

.438 

2.05 

20 

.342 

.940 

.364 

65 

.906 

.423 

2.14 

21 

.358 

.934 

.384 

66 

.914 

.407 

2.25 

22 

.375 

.927 

.404 

67 

.921 

.391 

2.36 

23 

.391 

.921 

.424 

68 

.927 

.375 

2.48 

24 

.407 

.914 

.445 

69 

.934 

.358 

2.61 

25 

.423 

.906 

.466 

70 

.940 

.342 

2.75 

26 

.438 

.898 

.488 

71 

.946 

.326 

2.90 

27 

.454 

.891 

.510 

72 

.951 

.309 

3.08 

28 

.469 

.883 

.532 

73 

.956 

.292 

3.27 

29 

.485 

.875 

.554 

74 

.961 

.276 

3.49 

30 

.500 

.866 

.577 

75 

.966 

.259 

3.73 

31 

.515 

.857 

.601 

76 

.970 

.242 

4.01 

32 

.530 

.848 

.625 

77 

.974 

.225 

4.33 

33 

.545 

.839 

.649 

78 

.978 

.208 

4.70 

34 

.559 

.829 

.675 

79 

.982 

.191 

5.14 

35 

.574 

.819 

.700 

80 

.985 

.174 

5.67 

36  ' 

.588 

.809 

.727 

81 

.988 

.156 

6.31 

37 

.602 

.799 

.754 

82 

.990 

139 

7.12 

38 

.616 

.788 

.781 

83 

.993 

.122 

8.14 

39 

.629 

.777 

.810 

84 

.995 

.105 

9.51 

40 

.643 

.766 

.839 

85 

.996 

.087 

11.43 

41 

.656 

.755 

.869 

86 

.998 

.070 

14.30 

42 

.669 

.743 

.900 

87 

.999 

.052 

19.08 

43 

.682 

.731 

.933 

88 

.999 

.035 

28.64 

44 

.695 

.719 

.966 

89 

.999 

.017 

57.28 

45 

.707 

.707 

1.000 

93 

1.000 

.000 

Infinity 

INDEX 


Absolute  units  of  force..  .   151 

Acceleration 112 

defined 113 

diagrams 153 

equations 117 

of  gravity 139 

of  rotation 164 

uniform 112 

Action ,    1,  5,  8 

Ampere 179 

Angular  velocity 109 

Aneroid  barometer 291 

Arch 105 

Archimedes'  principle  .  .  .   285 
Atmospheric  pressure  ...     89 

Atwood's  machine 145 

Average  velocity 115 

Axes,  coordinate  ....   36,  305 
Axis,  neutral 255-259 


B 

Barometer 290 

Beams,  strength  of 265 

shape  of 266 

Bending 254 

Bending  moment 258 

Bent  arm  lever 65 

Belt  machines 220 

friction  of 215 


PAGE 

Bicycle,  chain  drive 221 

Block  and  tackle 228 

Boyles'  law *0f  *f  I 

Brackets 267 

Brake,  Prony 215 

Bridge  truss 96 


Cast  iron,  strength  of  ...  267 
Center  of  gravity.  55,  58,  259 

of  pressure 281 

Centrifugal  force 171 

Centripetal  force 171 

Chain  hoist 228 

Charles'  law 294 

Circle,  motion  in 131 

Coefficient  of  friction  ....  199 

of  elasticity 240 

Collision 197 

Column,  shape  of 266 

Composition  of  forces,27,29,36 

of  velocities 25 

Compression 6 

Conservation  of  energy  .  .  208 

of  momentum 198 

Coordinate  axes 36,  305 

Constants,  table  of 299 

Couples 54 

Crane 88 

Curve  plotting 305 

317 


318 


INDEX 


Dalton'slaw 294 

Density 271 

Derrick 88 

car 22 

Diagrams,  strain 243 

work 176 

Differential  pulley 228 

Displacement 133 

Distance 115 

Ductility 245 

Dynamometers 214 

Dyne - 148 

E 

Efficiency 212 

Elastic  limit 241 

material 236 

strength.. 242 

Elasticity 5,  236,  244 

defined 238 

modulus  of  by  bending  263 

of  gases 271 

of  liquids 271 

of  volume  of  solids  ....  271 

Elevation  of  projectiles  ..  128 

Emerson  Power  Scale  ...  217 

Energy 181 

conservation  of  ...    182,  208 

kinetic 182 

of  rotating  bodies  ....  187 

potential 182 

transportation  of 183 

units  of 182 

Equation  of  straight  line.  309 

Equilibrium,  applied  ....  42 

conditions  of 50,  62 

defined 40 

Equivalent  weight 191 

Erg 176 


Factor  of  safety 243,  301 

Falling  bodies 121 


PAGE 

Flywheels,  speed  of 167 

Fluids 271 

Foot  pound 176 

Force 1 

centrifugal 171 

centripetal 171 

components  of 33 

defined 4 

graphical      representa- 
tion        16 

moment  of 12-13 

units  of 147 

Forces  at  the  joints  of  a 

structure 43 

composition  of 27 

effects  of 3 

measurement  of 4 

parallel 50 

parallelogram  of 44 

application  of  to  sus- 
pended bodies  . .     62 
application      of      to 

simple  truss  ....     45 

polygon  of 67 

resolution  of 33,  34,  35 

triangle  of 66 

Free  body 20 

Friction 5,  198 

coefficient  of 199 

laws  of 202 

of  lubricated  surfaces..   203 

rolling 203 

sliding 199 

static 199 

G 

Gases , . . .  271-  289 

Gearing 220 

Graphical    representation 
of  velocity,  forces,  etc.     16 

of  work  done 177 

Graphical  statics 148 

Gravitational  units 149 

Gravity 121 


INDEX 


319 


PAGE 

Gravity,  acceleration  of . .  139 

center  of 55,  58,  259 

specific 72 

Gyration,  radius  of 166 

H 

Harmonic  motion 131 

Heat,   mechanical  equiv- 
alent of 301 

Height  of  barometer  ....  292 

Hoisting  crane 178 

Hooks 267 

Hooke's  law 240 

Horse-power 178 

Hydraulic  press 277 

Hydrometers 286 


Inclined  pendulum 137 

plane 47,  153 

Indicator  card 177 

Inertia 5 

Input 211 

Iron,  strength  of 267 


Jackscrew 224 

Joule 176 

K 

Kilowatt 179 

Kinetic  energy 182 


Law,  Boyles' 293 

Charles' 294 

Dalton;s 294 

Hoake's 240 

Marriotte's 293 

of       conservation       of 

energy 208 

of  falling  bodies 121 


PAGE 

Laws  of  bending 254 

of  friction 202 

of  pendulum 136 

of  torsion 250 

Lever,  bent,  arm 65 

Liquids 271-289 

Limit,  elastic 241 

Lubricated  surfaces,  fric- 
tion of 203 

M 

Machines 210 

efficiency  of 212 

Manometer 296 

Mass 144,  147,  151 

Mechanical  advantage  ...   213 
equivalent  of  heat ....   301 

Mercury  barometer 191 

Modulus  of  elasticity  ....  240 

by  bending 263 

of  rigidity 251 

Moment,  bending 258 

Moment  of  a  force 12,  13 

of  inertia 166 

sign  of 50 

Momentum 196 

Motions 11,  108 

simple  harmonic 131 

uniform Ill 

uniformly  accelerated  .    112 

N 

Neutral  axis 255,  259 

Non-concurrent  forces  ...     81 

O 

Origin  of  coordinate  axes  305 
Output 211 


Parallel  forces 52 

Parallelogram  of  forces . .     27 
of  velocities 26 


320 


INDEX 


PAGE 

Pascal's  principle 276 

Passive  reactions 5 

Path  of  projectiles 128 

Pendulum 136-140 

inclined 137 

simple 136 

torsion 151 

Phase 135 

Pillar  crane 177 

Plotting  curves 305 

Polygon  of  forces 67 

Post  crane 77 

Potential  energy 182 

Poundal 148 

Power 177 

Pressure 273 

atmospheric 289 

center  of 281 

exerted  by  fluid 274 

gauge 296 

measurement  of  gas  .  . .   298 

standard 292 

Projectiles 128 

Prony  brake 215 

Pulley 228 

R 
Radian 110 

Radius  of  gyration 166 

Rate  of  change  of  velocity  113 

of  doing  work 177 

of  motion 108 

Reaction 1,  5,  8 

passive 5 

Resolution  of  forces,  33,  34,  35 

of  parallel  forces 52 

Resultant  velocity 25 

Rigidity,  modulus  of 251 

Roller  bearing 203 

Roof  truss 101 

Rotation 11 

S 

Safety,  factor  of 243 

Screw,  jack 20,  224 


Shaft,  speed  of 167 

Shear,  legs 247 

in  beams 259 

simple 91 

Significant  figures 302 

Simple  harmonic  motion.    131 

pendulum 136 

Sine  curve 135 

Solids 271 

Specific  gravity 272 

Speed 108 

of  fly  wheels 167 

of  shafts 167 

Sprocket  machines 220 

Squared  paper,  use  of. ...     35 

Stability  of  wall 282 

Static  friction 199 

Starting  torque 203 

Steel,  strength  of 267 

Steel     wire,     experiment 

with 236 

Stiffness 263 

Stone  tongs 95 

Straight  line,  equation  of  309 

Strain 239 

diagram 243 

Strength 244 

elastic 242 

of  beams 265 

ultimate 243 

Stress 238 

for  hook,  bracket 267 

general  equation  for. . .   260 
maximum  compressive.  256 

tensile 256 

Struts,  strength  of 266 

Sulkey  derrick 103 


Table  trigonometric  func- 
tions    316 

Tank,  water 104 

Temperature,     effect    on 

gases 289 

standard 289 


INDEX 


321 


PAGE 

Tension 6 

in  hoisting  ropes 157 

Torque 15 

Torsion 249 

laws  of 250 

pendulum 251 

Transmission    dynamom- 
eters     216 

Transformation  of  energy  183 

Translation 11 

Triangle  of  forces 66 

Trigonometric  functions, 

66,  313 

Truss 96 

or  roof 7,  101 

Abridge 96 

Twisting  moment 251 


U 

Units,  absolute 151 

gravitational 151 

of  force 153 

of  mass 153 

of  power 178 


PAGE 

Ultimate  strength 243 

Uniform  motion  .  .111 


Variable  motion Ill 

Velocity 12,  108 

angular 110 

average 115 

graphically  represented     16 

parallelogram  of 26 

Velocity  ratio 212 

Velocities,  composition  of     25 
Volt 179 

W 

Wall  crane 84 

Water  tank 104 

Watt 179 

Weight 151 

Weber  dynamometer  ....   217 
Weston  differential  pulley  228 

Winch 219 

Wrought  iron,  strength  of  267 

Work 175 

diagrams 176 

units.  .  .176 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


09824 


M282536 


QCiZS". 

0*3 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


